 In the last lecture, we saw equations of fluid motion, namely the law of mass conservation applied to the bulk fluid as well as the second law of motion also called the Navier Stokes equations. Moving fluids also carry with them scalar quantities. Today, we are going to look at equations that govern transport of scalar quantities. The first scalar quantity is the species in a mixture. For example, if you are dealing with a combustion problem, then the species carried are oxygen, carbon dioxide, fuel, carbon monoxide and so on and so forth. The law of mass conservation for a species in a mixture is called the mass transfer equation. We will also invoke the first law of thermodynamics which transports energy in a moving fluid. Both energy and the concentration of a species are scalar quantities. So, what does the law say for a species? For a species k in a mixture, the rate of accumulation of mass of the species k that is given by m dot k accumulation equals rate of mass in minus rate of mass out plus rate of generation of species within the control volume. As a result of chemical reaction, as you know, some species are generated and while others are destroyed. As such, each species will have either a generation or a destruction rate associated with. So, what is m dot k accumulation? That is simply the mass of the species k within the control volume delta v d by dt of rho k delta v. The mass of species in will be the flux of species 1, the flux of species k in direction 1 multiplied by the area d a 1 which is this area plus n 2 k which is coming from the bottom multiplied by the area delta a 2 plus n 3 k multiplied by delta a 3 at x 3. Likewise, the same quantities at the outgoing phases at x plus delta x 1, x 2 plus delta x 2 and x 3 plus delta x 3 rho k is the species density. If we now substitute these expressions in this verbal statement and divide each term by delta v which is the product of delta x 1, delta x 2, delta x 3 and let each of these increments tend to 0. This is a procedure we have gone through before. Then you will see, you will get an equation of this type d rho k by dt plus net transport of species k in direction 1 plus net transport in direction 2 plus net transport in direction 3 equal to rate of generation of species k. Now, the total mass flux n i k in direction i is the sum of the convective flux due to rho k u i due to bulk fluid motion and diffusion flux m double prime i k due to density difference. Thus, n i k is represented as rho k u i plus m double dot i k. Writing convective flux in this manner is indicative of the fact that we are assuming that each species is travelling or is being carried at the same velocity as the bulk fluid. The diffusion flux on the other hand m double dot i k arises simply due to the differences in density at neighbouring locations. The expression for mass flux by diffusion is given as m double prime i k equal to minus d rho k by dx i. This equation is analogous to the conduction heat transfer and is like a Fourier's law of heat conduction. In mass transfer, it is called the Fick's law of mass diffusion. Just as in convective heat transfer, you know that the total flux of energy is given by the convective flux plus conduction flux. In mass transfer, we say it is the convective flux plus diffusion flux. d is called the mass diffusivity. It has units of meter square per second. If I were to substitute for n i k for each of these terms, then I can rewrite this equation in the following manner. It would read as d rho k by dt plus d rho k u 1 by dx 1 and so on so forth plus is equal to d by dx 1 of d rho k by dx 1 and likewise in direction 2 and 3 plus r k. Rho k has units of density. It is customary to define mass fraction omega k as the specific density divided by the mixture density. Therefore, some sigma omega k equals 1 by Dalton's law. Another way of saying is sum of densities is equal to the mixture density. The notion behind Dalton's law is that each specie behaves as though it occupies the volume of the total mixture. So, if I were to substitute now rho k as rho m multiplied by omega k, this equation would be written in tensor notation in this fashion. This is called the mass transfer equation. It has a transient term, a convection term, a diffusion term and a source or a generation term. Now, let us sum each of these terms over all species. That is sum over all k as in this as done here. Then clearly here, this term would reduce to d rho m by dt. This term would reduce to d rho m u j by dx j because omega k is equal to 1. Sigma omega k will simply become unity or constant and therefore, this term would simply vanish. Just see now that this equation would be d rho m by dt plus d rho m u j by dx j equal to sigma r k and that term is 0. You will recognize readily then in the absence of omega k, the left hand side is simply the bulk mass conservation and it equals 0. It follows therefore, that sigma r k must be 0. These two last deductions are very important. Sigma r k equal to 0 says that whenever there is a chemical reaction, it is true that some species will be generated but there will be others that would be destroyed and the total mass cannot be generated or destroyed. That idea is expressed in sigma r k equal to 0. What does summation of diffusion equal to 0 imply? It implies that when some species are being diffused in a certain direction, other species are being diffused in the opposite direction and that stands to reason. For example, if fuel was decreasing, then product would increase. The net result is that summation of all diffusive quantities would be 0 and therefore, sigma n i k would be sigma rho m u i plus sigma m double prime i k. That quantity is 0 and therefore, some of the mass fluxes in direction i are simply the bulk mass flux rho m u i. We shall refer to this equation much later in the course, a word about mass diffusivity. Strictly speaking, mass diffusivity is defined only for a binary mixture of two fluids 1 and 2 as D 1 2. But in a combusting product mixture, for example, there are several species present and diffusivities of pairs of species are true to the mass diffusivity. So, diffusion of carbon dioxide in nitrogen or the diffusivity of carbon dioxide in nitrogen is different from diffusivity of oxygen in nitrogen and vice versa. But in gaseous mixtures, these diffusivity pairs between species tend to be very nearly equal and therefore, we discard D 1 2 or D i j as the symbol for diffusivity and replace it by a single symbol D and that suffices for combustion calculations. Incidentally, in turbulent flow, this assumption of equal effective diffusivity holds even greater validity. This we shall appreciate a little later. We now turn to the first law of thermodynamics. In rate form that is watts per cubic meter, the first law of thermodynamics reads as D e by D t equal to D q by convection by D t, D q conduction by D t plus rate of generation of energy minus work done by shear forces minus work done by body forces. Each of these terms is defined here. We shall seek mathematical representations for each one of them. So, let us consider the first term E dot. That would simply be D by D t rho m E naught or the total energy V by D t. The total energy is the sum of static energy plus kinetic energy and the static energy by thermodynamic relation is nothing but enthalpy minus p into specific volume or p divided by rho m plus V square by 2. So, E m is mixture specific energy, H m is mixture specific enthalpy, V square is the kinetic energy, sum of u 1 square, u 2 square and u 3 square and rho m as we saw before is sigma rho k the mixture density. In general, E naught will have contributions from many other effects like potential energy associated with rise of all in elevation, electromagnetic energies and so on and so forth. But these we will neglect because practical equipments are fairly small, but in which kinetic energy would be of some interest, but not these energies that I mentioned at the moment. Now, in order to represent the heat transfer and the work transfer across a control volume, we shall follow the thermodynamic convection. We will say that heat flow into the control volume is positive whereas, the heat flow out of the control volume is negative and as we said before, all species are transported at mixture velocity. So, here I have shown all the net mass transfers and heat transfer that would take place across a control volume phase. N is total mass flux and Q is the total heat flux across the interface, across is control volume phase. So, net convection then by following the convection that the heat in is positive whereas, heat out is negative, we would have convection is simply d by dxj of sigma n j k E naught k and if I replace E naught k by h k plus p by minus p by rho m plus v square by 2, then you will see n j k would multiply with h k into the bracket. But the sum of n j k would simply be rho m u j as we saw before minus p by rho m plus v square by 2. If we now note that omega k into h k that is the mass fraction of specific k multiplied by its specific enthalpy must add up to the mixture enthalpy, then sigma n j k h k would be sigma rho m u j omega k plus m double dot j k the diffusion flux into h k and since sigma omega k h k is equal to h m, it will add up to rho m u j h m plus sigma m double prime j k h k. Therefore, if I replace this term here, then you will see you have rho m u j h m minus p by rho m plus v square by 2 which would be written in this fashion and then the leftover term the diffusion flux term that that would be given by that term. So, that is the expression for the convective flux. Net conduction again by Fourier's law of conduction q conduction will be simply minus q j by d x j where q j is k m multiplied by d t d x j and k m is the mixture conductivity net volumetric generation. Now volumetric generation in a moving fluid typically would comprise of the generation due to chemical energy because some reactions are exothermic, whereas some other chemical reactions are endothermic. So, q dot k m would be positive in case of exothermic reactions and it would be negative in terms of endothermic reaction. So, if you want to evaluate q dot k m in a moving fluid, then one needs to postulate first of all the chemical reaction model that is being employed and there are variety of chemical reaction mechanisms of different levels of complexity. We shall see all this when we come to study of mass transfer. q rad represents the net radiation exchange between the control volume and its surroundings. Now usually evaluation of this term would require a what is called the radiation transfer equation and it happens to be an integral differential equations. Treatment of that equation is really beyond the scope of the present lectures and it would require usually numerical calculations in a real practical equipment. But suppose the mixture that we are carrying has very high absorptivity may be because of suit, may be because of particulates that are present in the gaseous mixture, then the absorptivity and scattering of the radiation would be very high. So, when absorptivity and scattering coefficients are large, q dot rad can actually be represented in a manner similar to the conduction equation and the radiation conductivity can be defined as 16 into sigma t cube divided by a plus a's where sigma is the Stefan-Boltzmann constant that you are all familiar with. But remember this kind of representation is justified only when the gaseous mixture has very high absorption and scattering coefficients. Now we come to the work done terms. Now the derivation to follow is somewhat wrongish and I would request you to play good attention to how the derivation progresses. So, firstly there are shear forces and normal forces. So, here is a stress. Stress is force per unit area multiplied by velocity gives you the work done and d by dx 1 of that gives you the net work done. So, sigma 1 which acts in the direction 1 multiplied by u 1 plus tau 1 2 into u 2 is the work is the shear work in net shear work in external direction and so on and so forth. So, you will get shear stress multiplied by the associated velocity in the two directions and the net work done by stresses w dot s would be given as that. The body forces work is rho m into b 1 is the force multiplied by the velocity u 1 is the work done in direction 1, 2 and 3 likewise. So, w dot s is the stress work, w dot b is the body force work and now we shall write these things as differentiation of a product. So, d by dx 1 sigma 1 u 1 will be written as sigma 1 d u 1 by dx 1 plus u 1 into u d sigma 1 by dx 1 and so on and so forth and on the next slide you will see the result. So, the total work done will be u 1 multiplied by all these terms u 2 multiplied by all these terms u 3 multiplied by all these terms plus sigma 1 d u 1 by dx 1 sigma 2 d u 2 by dx 2 sigma 3 d u 3 by dx 3 tau 1 2 into the strain rate associated with tau 1 2 tau 1 3 into the strain rate associated with tau 1 3 and tau 2 3 is the strain rate associated with tau 2 3 and here I have already used the idea of complementarity of stresses that is tau 1 2 is equal to tau 2 1. Now, I draw your attention to this term. If you recall when we wrote the Navier-Stokes equations before or the Newton's second law of motion these are simply the net forces acting in direction 1 and therefore, these are nothing but the right hand sides of the Newton's second law of motion. They can be replaced by the left hand side of the Newton's second law of motion and that is what I will do on the next slide plus all the stresses can be replaced in terms of viscosity multiplied by strain rates. So, let us do this first. So, multipliers of u 1, u 2, u 3 in equations 89 are simply right hand sides of momentum equations. If you recall that on lecture 3, slide 13, 14, 15 we have replaced them by left hand side of momentum equations. So, left hand side was u 1 d u 1 by d t into rho m u 2 d u 2 by d t and u 3 d u 3 by d t and this would simply be d u 1 square by 2 d t d u 2 square by d t and d u 3 square by d t which we write as d v square by 2 d by d t into rho m. So, that is as these are the first three terms that we have expressed. Now, we look at the remaining terms. So, if I replace sigma 1 by minus p plus tau 1 1 sigma 2 as by minus p plus tau 1 1 sigma 2 as minus p plus sigma plus tau 2 2 and sigma 3 as minus p plus tau 3 3 and tau 1 2 as mu times d u 1 by d x 2 into d u 2 by d x 2 and so on and so forth. You will see that that term would be equations 21 and 22 would be simply mu phi v minus p del dot v where phi v is called the viscous dissipation function and it would take the form 2 times d u 1 by d x 1 square which arises from tau 1 1 d u 2 by d x 2 square d u 3 by d x 3 square and so on and so forth. Del dot v as you recall is simply d u 1 by d x 1 plus d u 2 by d x 2 plus d x 3 by d x 3 and therefore, the total work done terms are simply rho m into d by d t v square by 2 plus mu phi v minus p dot v. You will notice that phi v will always be positive because it is a sum of all the gradient squared. So, that term would always be positive and in general its purpose is to increase the energy level of the mixture. This is the kinetic energy term and this is the minus p d v term or sometimes also called the pressure work term. Del dot v as you know is this. By way of summary, then we wrote