 So, now we will look at the Reynolds flow model which is an algebraic model as we said earlier unlike the Stefan flow model which is a differential equation model, but a one dimensional differential equation model and the Couette flow model which again was a one dimensional model. Our interest now is to develop the features of the Reynolds flow model and what I will do is first of all define Reynolds flow model again go over the Reynolds flux hypothesis Reynolds flow model is supposed to mimic the boundary layer flow model we will also consider momentum transfer with suction and blowing. We will also consider single phase convective heat transfer with suction and blowing and then the four types of mass transfer problems in which the species mass transfer equation as well as the energy equation will be involved. I am now going to define the Reynolds flow model as you recall this is the interface this is the considered phase to which the mass transfer N w is taking place. This is the transfer substance phase from where the mass transfer is taking place to the considered phase and this is the infinity state which is far away from the interface in the considered phase. So, the principle idea of the Reynolds flow model revolves around postulation of a fictitious flux. It is a notional flux kilograms per meter square second. So, it is a notional mass flux define in the infinity state and its direction is towards the interface and in doing so as it flows from the infinity state towards the interface or the w state it carries with it the properties of the infinity state. So, that is the first postulate of the model and the model develops algebraic relations that are related to the real transport phenomena by reference to this flux G and further the model will result into N w equal to G times b. So, instead of the logarithmic relations we saw earlier the model will develop N w equal to G times b as the appropriate relationship. Now, the objective of course, in order to solve the problem we first need to define b of course, which we have done in the previous two lectures for a variety of mass transfer problems and the second thing is to evaluate G, but the way we will do it is our root will be that G will be related to the value of the heat transfer coefficient in the absence of mass transfer. So, we are going to relate G to heat transfer coefficient for V w equal to 0 and the reason we do this is because these values are available for wide variety of situations through correlations. So, given a mass transfer situation let us say flow over a cylinder like this then I would know what the Nusselt number is as a function of Reynolds number and Prandtl number. Nusselt number based on diameter d I would know what it will be as a function of Reynolds number and Prandtl number and for which I can divide H of V w equal to 0. If I can relate this value to G which is in kilograms per meter square second whereas, H is in watts per meter square or essentially joules per meter square second. So, obviously we expect a relationship between G will be proportional to H cough V w equal to 0 divided by specific heat. Now, this will have the same units because H cough is joules per meter square second and C p is joules per K G per Kelvin divided by meter square Kelvin K divided by K divided by K. So, K K gets cancelled joules and joules get cancelled and again you will get K G per meter square second. So, this is really this is what we are aiming at this is what we are aiming at. What therefore, this will do is the following that you given a mass transfer situation given a mass transfer situation of any kind this can be a mass transfer flow over a cylinder or it can be flow through a tube or anything like that we calculate the appropriate value of V and evaluate G from the knowledge of the heat transfer coefficient in the absence of mass transfer in that same situation. So, basically no differential equations are solved the entire model is algebraic. Now, in order to facilitate further discussion I am going to define in the W state two more surfaces which are almost coincident with the W interface. One of them is the T W state which is just inside the considered phase and T L state which will be just inside the transfer substance phase. So, this is what T W and T L are the T in the T state as you remember there are no temperature or composition gradients and they are uniform. In addition we shall also postulate that there will be conduction heat transfer from the considered phase to the W state and there can also be possibly some temperature variation from deep inside the transfer substance phase to the interface in which case there would be some conduction heat transfer even in the transfer substance phase which I have called Q L this is Q W in the considered phase. There can also be radiation heat transfer to the wall in the considered phase and in this model for convenience we will account for it as though the radiation flux actually penetrates the interface and appears in the transfer substance phase. The reason for doing this would be very obvious as we go along. So, there are few new definitions introduced T W state, T L state, heat flux Q W and heat flux Q L in the considered and transfer substance phases respectively and the Q radiation which is accounted in the transfer substance phase. The mass transfer flux is the sum of all species transferred across the interface. So, N W can be sum of N J W let us say in the transfer substance phase and N K W in the considered phase. The species J in the transfer substance phase need not be the same as species J, species K in the considered phase. This is typically the case when let us say I have a mass transfer taking place across a wood particle then here the wood itself has a formula of C X H Y S I mean sorry H O P S Q and so on and so forth. And when the wood burns for example, you will get all sorts of gases like CO 2 CO or H 2 depends on the temperature of the surface. So, the species involved inside the wood surface need not be the same as involved in the considered phase, but the sum of those fluxes of these species and the sum of the flux of these species should be the same. We say Q W is positive when flowing towards the interface from the considered phase and Q L is positive when flowing away from the interface in the neighboring phase. So, in other words in this figure both Q W and Q L are shown in the positive direction. If I now draw an interface say W W and let us say this is T W and this is T L there is Q W here and Q sorry Q L in this direction. Then the mass transfer flux is passing like this N W. So, if I take the control volume like this, then N W into H T L which is coming in plus Q W which is also coming in must equal what is going out which is Q L plus N W into H T W. So, N W T N W T N W T N W into H T L comes in and what is going out is N W H T W that is what we mean. Therefore, you will see that Q W minus Q L would be equal to N W into H T W minus H T L. Therefore, Q W minus Q L will be N W times H T W minus H T L where H T W and H T L are enthalpies of the transfer substance at the T W and T L state. This is very important. H T W simply implies the enthalpy of the transfer substance if it was to appear in the considered phase as H T W. H T L on the other hand is the enthalpy of the transfer substance in the neighboring phase, but at the temperature of the interface. So, that is what we mean by enthalpy here. So, suffix T W and T L are different from suffix W itself. Suffix W refers to the mixture enthalpy whereas, suffix T W refers to the transfer substance enthalpy both in the W and L states. That is just inside the considered phase and inside the transfer substance. Now, if Q W minus Q L is not equal to 0 then a phase change occurs. For example, if H T W minus H T L was equal to the remember H T W is the vapor phase. For example, in an evaporation problem and H T L will be the liquid enthalpy then H T W minus H T L would be simply H F G the latent heat of evaporation. Therefore, Q W minus Q L will not be equal to 0 and that is where when the phase change occurs. So, that is the idea of why we define H T W and H T L as the enthalpies of the transfer substance in the vapor in the considered phase and in the transfer substance phase. Q rad from the considered phase to the interface is positive it is accounted in the neighboring phase. So, with these definitions we now turn to the main hypothesis of the model. Now, in different applications with or without chemical reaction complex processes occur in the considered phase that is where you will have effects of property variations effect of turbulence and variety of other there may be chemical reaction there may not be any chemical reaction so on and so forth. This sort of situation would require solution of the complete set of boundary layer equations and what the algebraic model or the Reynolds flow model tries to do is to say that the to avoid solution of any differential equations it is assumed that the complex processes can be effectively simulated by a fictitious flux G through the infinity infinity state towards the interface carrying with it properties of the infinity state. In other words G flux carries with it the properties of the infinity state towards the interface. On the other hand N W plus G flux carries with it is again postulated at the infinity state, but it carries with it values of the W state values of the upstream values if you like likewise upstream values here in this direction. This is a very important distinction between the two fluxes defined that G flux carries with it the properties of the infinity state as it moves towards the towards the interface whereas N W plus G moves away from the interface and carries with it the values of the W state. That is what I have said a fictitious flux N W plus G through infinity infinity plane away from the interface carrying with it properties of the W state. What this ensures is that in the considered phase there will be no net generation of mass at all because G is coming in and G is going out. So, there is no net generation of mass and that is what is expected even in a real boundary layer flow there should be no net generation of mass. The magnitude of G is not affected by the presence of gradients of omega j t or turbulence radiation etcetera in the considered phase. So, that is another characteristic of the G flux. It is also not affected by the direction of N W in other words I may have a problem of evaporation in which N W is this way I may have a problem in which condensation occurs in which the N W is negative. So, G is not influence by presence of by the direction or magnitude of N W and secondly it is not affected by gradients of omega j temperature or turbulence quantities in the considered phase. And yet in spite of these assumptions made G flux is supposed to produce the same effect at the interface that the real boundary layer flow is likely to produce. This is the claim of the model this is the claim of the hypothesis that it can in fact represent all effects of a real boundary layer flow although the method is very simple and algebraic. To illustrate this let me consider the case of evaporative cooling. So, for example let me say that I have a very hot surface here at temperature T s and it is hot because very hot gases are flowing over it in the infinity state very hot gases are flowing and they are dry very little humidity. Obviously, to protect this surface from the hot gases and to keep its temperature down what one might do is to provide let us say a water film here pass a thin film of water. Then what it will do is that the water film will evaporate the water film will evaporate and in the process will receive heat both from the solid surface as well as from the hot gases. So, this is this is the likely temperature profile. Remember in the absence of the film probably the temperature will have been much higher than what is shown here. So, in order to cool down but at the liquid film the temperature will drop considerably and that would result into actually some transfer of heat from through the liquid film from the wall to the film to the age of the film and likewise there would be heat transfer. So, q w in this case would be positive, but q l which would be now in the from the transfer substance phase to the interface would be negative q l would be negative and n w would be positive. So, what we expect therefore, the Reynolds flux hypothesis states that a g flux of hot gases towards the interface together with the n w plus g fluxes of cooler moist air away from the interface away from the interface hot gases towards interface will produce the effect that n w is greater than 0 q w is also greater than 0 because it flows into the surface and q l will be negative because it is also flowing into the surface from the neighboring phase. Now, this is what we expect the real flow to do and the Reynolds flux model claims that it will do precisely that that it will do precisely that. So, in this sense the g flux is considered capable of responding to the mass fraction and temperature gradients as well as turbulence effects in the considered phase as shown in this on this side. So, it disregards what the temperature gradients here are what the temperature gradients here are and so on so forth or the mass fraction gradients here are we can remember mass fraction gradient would fall. It disregards any turbulence effects in the considered phase and will produce exactly the same effect n w greater than 0 q w greater than 0 and q l less than 0. That is the claim of the model. Let us see how we can justify that because it can handle both momentum and heat transfer. We are going to consider the control volume between infinities and w states. So, basically what I am going to do now is to consider the w state and infinity state and I am going to consider momentum transfer. Then the g flux will bring with it g times u infinity. So, momentum in will be g times u infinity momentum out will be n w plus g u w n w plus g u at w plus shear stress tau w that is also momentum out of the cell. But at the surface u w will be 0. Therefore, that goes out. So, basically if I equate the two I get g u infinity equal to tau w that is what the Reynolds flow model implies. So, let us see what does this mean? Well, so basically as you will see here I have said n w plus g u w equal to 0. Therefore, g u infinity divided by tau w all would equal 1 and n w divided by rho times v w. So, what I have done here is 1 is equal to n w itself is rho v w. So, I have replacing 1 by n w by v w or n w can be written as g times rho v w u infinity by tau wall that would be equal to g into v w divided by u infinity tau wall over rho u infinity square. Essentially I have divided and multiplied by u infinity and what is this quantity? This is essentially v w by u infinity divided by c f x by 2 and therefore, n w is simply g times v f. Our claim that the model is able to predict all effects of momentum transfer through this expression n w equal to g b is verified as long as we interpret b in this fashion v f equal to v w u infinity divided by c f x 2. So, in order to get g what will be g? The meaning of g will be tau wall over rho u infinity which will be rho u infinity c f x by 2. So, in the momentum transfer problem g would be simply rho u infinity c f x by 2 and c f x by 2 is available to us from correlations for many problems. Let us look at single phase convective heat transfer in which the consider the c v between t and infinity states. Now, what I am going to do is to consider this is the w state, this is the t state and this is the infinity infinity state. Then you will see that heat in will be g times h infinity, there will be n plus w n w plus g h w and there would be here n w times h t. This is the so I am considering the control volume between infinity and infinity and t t states. So, there is a it will bring with it g u h infinity as the enthalpy total energy flux it there will be n w into h t will be the total energy flux coming in here also and here and what will be going out is n w plus g h w and that is going out. If I quit what is coming in and what is going out then you will see that I get this relationship rate of heat flux in is equal to g h infinity plus n w h t plus n w plus g divided by multiplied by h w. So, if I quit and rearrange then again you will see I get n w equal to g times b h where b h is equal to h infinity minus h w divided by h w minus h t or h infinity in this case because a single phase mass transfer a single phase convective heat transfer with suction and blowing maybe this will be simply c p into t in the infinity state minus c p into t into w state thereby c p t w state minus c p t in the transfer substance phase and therefore, n w will be equal to n w v w. If the specific heats are equal of course b h would be simply t infinity minus t w divided by t w minus t t if I now consider infinity infinity state n w states. So, if I now consider those states then you will see here you will get n w h w. So, my equations would now read as g h infinity coming in n w h w coming in and n w plus g h w going out. So, that is what I have written plus there would be q w also which will go out between in this control volume or this control volume. So, you will see that I get rate of heat flux in as g h infinity plus n w h w rate of heat flux out will be n w g h w plus q w and if I equate and rearrange then I will get g is equal to h times q w divided by h infinity minus h w and if specific heats are equal in the infinity n w states then g would be simply q wall divided by c p t infinity minus t w and as you know this would be simply minus h cough v w by c p. So, therefore, g is indeed related to heat transfer coefficient in the presence of suction and blowing. So, that is what the single phase convective heat transfer shows convective momentum transfer shows that g is equal to rho u infinity c f x by 2 and here h cough is equal to. Now, you can see very well that if I define if I since g is equal to minus h cough v w divided by c p if I divide and multiply this by rho u infinity rho u infinity then you will see this will be stanton x into rho u infinity. So, rho u infinity stanton x for v w analogous to rho u infinity c f x by 2 that was from momentum transfer and this is for heat transfer very similar formulas which you essentially implies that the analogy between heat transfer and momentum transfer through the coefficient g the convective mass transfer coefficient. So, inert mass transfer without heat transfer. So, now we consider now let us say in all states the temperatures are equal. So, there is no heat transfer at all, but there would be mass transfer. So, again if I consider t control volume between t v t state and infinity state then there will be rate of mass flux in will be g v omega v infinity plus n w into omega v t which is from the neighboring phase in the transfer suction state and what will go out would be n w plus g omega v w because the outgoing flux carries with it properties of the w state. Again if I rearrange equate in and out then you will get n w equal to g v m where v m now is omega v infinity minus omega v w omega v w minus omega v t and for a pure liquid in the t state omega v t will be equal to 1 and how would we evaluate omega v w it would be evaluated from equilibrium relation at t w. So, this is what Reynolds flow model shows even for inert mass transfer without heat transfer now let us consider inert mass transfer with heat transfer. Now in this case let us say infinity state is not temperature is not equal to t w and not equal to t t then the c v between infinity and t states will show that the rate of heat flux in will be g times h m infinity plus n w h m t and what would go out will be n w plus g into h m w again equating you get n w equal to g v m h where v m h is equal to h m infinity minus h m w divided by h m w minus h m t. In an inert mass transfer case let us say air and water vapor mixture then h m would be omega v times enthalpy of the vapor and omega a times which is the mass fraction of a into enthalpy of air and since there are only two spaces omega a air would be written as 1 minus omega v and enthalpy of the vapor would be c p v t minus t ref plus lambda ref which is the latent heat defined at t ref and h a would have simply sensible heat c p a t minus t ref. As you recall if we make the Lewis number equal to one assumption then b m h would be equal to b m from which omega v h w t w relation is iteratively calculated so we can since n w is equal to g b m h and n w is equal to g b m so b m is equal to b m h and we can calculate the in case the unknown t w is equal to b m h and omega w states can be evaluated from the equilibrium relationship iteratively. This is exactly what we did even in Stefan flow and Couette flow models. Now a few more interesting results if I consider the c v between w state infinity state and w w state and right then you know this is q w this is here I will get h n w times h t w this will be g times h m infinity and there would be g times plus h m infinity and there plus n w into h m w. You can see now the difference between h m w and h t w h t w is the is the enthalpy of the substance which has been transferred from the neighboring phase into the and therefore if I energy in will be will be equal to g into h m w. So, g h m infinity plus n w h t w and energy flux out will be equal to g plus n w h m w plus q w plus q w and that is what I have written here. So, on the left hand side you have the fluxes going out and on the right hand side you have the fluxes going in and therefore if I now rearrange these equals equality then you will see B m h would now get defined as h m infinity minus h m w h m w minus h t w plus q wall by n w. But, you will recall we showed that q w minus q l where q l is the heat transfer in the transfer substance phase equal to n w t w minus h t l. So, I can replace q w from this relationship and I get another expression for B m h in terms of q liquid divided by n w. On the other hand if I also consider the radiation in the transfer substance phase then you will see that I get if I consider radiation in the transfer substance phase then this is the w w surface this is the t t surface. Then q l is saw q rad is saw this is n times h m t and here you have n w times h t l h t l. So, I would get essentially n w h m t plus q l plus q rad equal to n w h t l. So, I would get essentially n w h m t plus q l plus q rad equal to n w h t l h t l. So, I would get essentially n w h m t plus q l plus q rad equal to n w h t l h t l this would be the energy balance between w and t surfaces where in the previous presence of radiation and then you will see that I can even replace q l divided by n w h m t l by minus h m t minus q rad by n w. So, variety of expressions for B m h can be found when inert mass transfer with heat transfer. This is of great value relevance because different expressions come handy depending on the problem at hand or the mass transfer problem at hand as we will discover when we solve practical problems. We turn to mass transfer with heat transfer and simple chemical reaction. Now, as you know in this case 1 kg of fuel reacts with r s t kg of oxygen to produce 1 plus r s t kg of product with the where r s t is the kilograms of oxygen to a kilogram of fuel is the stoichiometric coefficient. So, I can write between infinity and t states an equation for the fuel. This is the amount of fuel mass flux in. This is the amount of fuel mass flux in from the bottom n w and this is the amount of fuel that will go out of the considered phase carrying with it value of omega f u w and all of that must equal minus r f u dot. For oxygen likewise same expression equal to minus r o 2 because in the considered phase in the presence of chemical reaction both fuel and oxygen would be depleted. So, that is why I have shown these are the area this is the volume averaged reaction rates r f u and r o 2. But r o 2 is equal to r s t r f u and therefore, if I divide this equation by r s t and subtract from the first equation I would again get essentially n w equal to g phi where b phi is phi infinity minus phi w and phi w minus phi t where phi is as you know is a conserved property omega f u minus omega o 2 by r s t. By the same reasoning I can also form and write an equation for product here and I would get phi equal to omega f u omega product plus 1 plus r s t or omega product 1 plus r s t plus omega o 2 by r s t. Now, these we have gone over this type of material before the important thing is even in the presence of chemical reaction we are able to show that n w equal to g b holes where b is now formed from appropriate conserved property. Here, we had invoked the mass conservation principle. Now, we will invoke the energy conservation principle and assume C p k equal to C p m as we have done earlier also. Then H m will be omega k H k C p m t minus t ref omega f u del h c and that I can also associate del h c with omega o 2 by r s t. So, if I consider C v between infinity and w states then g H m infinity plus n w H m t w would equal g plus n w H m w plus q w. If I take the second definition of H m then again we would have n w equal to g b m H where if I now take t ref equal to t w itself then b m H would be H m infinity minus H m w H m w minus H m t w plus q w n w divided by n w that would result into C p m t infinity minus t w plus del h c omega o 2 infinity minus omega o 2 w divided by r s t del h c omega o 2 w minus omega o 2 t w divided by r s t plus q w by n w. Now, of course, you can imagine supposing I was burning a liquid fuel it would have no oxygen. So, that would be sorry it would have no oxygen and therefore, that would be 0. If the liquid was at its boiling point then no oxygen can survive there and therefore, that would be 0 and I would get simply C p m t infinity minus t w del h c omega o 2 at infinity divided by r s t and for a liquid droplet burning in air I would readily know what omega o 2 infinity is. The only thing that I now need to do is to interpret q w. So, for a volatile fuel or transpiration cooling by combustible gas like hydrogen omega o 2 w is 0 and since the transfer substance does not contain oxygen I would get this expression that is what I showed. If the liquid fuel is at this boiling point t w is equal to t b p then q w minus q l would be equal to n w h f u and therefore, q w divided by n w will be simply C p m t infinity minus t b p del h c omega o 2 infinity divided by r s t lambda f u plus q l divided by n w. Now for an atomized tiny droplet we can say that there is hardly any temperature difference between w and t states. It is like saying that we have injected inside the diesel engine the liquid droplet already at its boiling point in which case there can be no conduction heat transfer in the transfer substance phase and therefore, you will see that with q l equal to 0 and t t equal to t b p you will get C p t infinity minus t d p delta h c omega 2 infinity by r s t plus lambda f u divided by lambda f u and that quantity would be 0. So you can see B m h can now be evaluated very easily because boiling point of a liquid is known delta h c the heat of combustion of the fuel is known omega o 2 infinity would be typically known and the stoichiometric coefficient in the simple chemical reaction would also be known along with the latent heat of the fuel. So all quantities are known and B m h can be evaluated very very easily. Now when it is difficult to ascertain mass fractions of compounds in different states it is preferred to use the conserved property n alpha equal to n alpha k omega k n alpha k being equal to the molecular weight of alpha the element alpha divided by molecular weight of the species in which it occurs. So again if I do C v between infinity and t states then element balance would give me n w equal to g by B m very easy to see this and B m would be again eta alpha in infinity w w and t states. Many times we will find that it is much more convenient to combine different eta alpha into a new conserved property variable phi as was shown in lecture number 32 where we considered the burning of graphite. In all cases we now find that the Reynolds law model results into n w equal to g B where B is equal to psi infinity minus psi w divided by psi w minus psi t psi can be in in only omega B it can be element fraction eta alpha it can be mixture enthalpy H m or it can be an appropriately defined phi from the participating species. Any linear combination of psi's are also solutions. So we can add any of these types of psi for our convenience. The result n w into g from the Reynolds law model is different from the result n w g star ln 1 plus B from 1 d Stefan flow model which was essentially diffusion model and the cured flow model both of them had logarithmic form whereas here it is it is a it is a g into B it is like much more like the Ohm's law type of model. The result from the model correctly identifies B F in momentum transfer and shows that g would be equal to H cop v w by C p in single phase convective heat transfer. So the Reynolds law model result will be employed to provide the interface boundary conditions to the boundary layer flow model in the next lecture and we will see how this can be done in the next lecture.