 In all previous lectures, our interest has been to consider heat and mass transfer close to a wall, whether force convection or natural convection. In this last lecture of this course, I am now going to consider a flow with heat mass transfer and chemical reaction in which no wall is present and that is in everyday language is nothing but the combustion flame. And my task is to see how we can apply the knowledge we have gained so far to predict the length and thickness of a diffusion flame. So, I will proceed as shown here. I will first of all define a flame and then I will set up the governing equations. Then first consider laminar jet whose velocity would be predicted and then flame length and shape would be predicted of the laminar jet flame. And then likewise, we will do the same in for the turbulent jet flame. Remember again that the word jet implies that there is no wall present and it sometimes also called a free shear flow. So, let us look at the definition of a flame. Here is the definition. The most commonly encountered diffusion flame is the candle flame that you can see here. In diffusion flame, the source of the fuel and the oxidizer are physically separated. So, the candle flame is an example in which the combustion of gaseous hydrocarbon fuels, sorry the candle flame is an example in which the melted wax here evaporates. And so, the fuel alone arrives into the flame where the oxygen required is gathered from the surroundings. So, fuel itself does not have any oxygen in it, but the oxygen required for burning is obtained from the surroundings by process of diffusion. Gaseous hydrocarbon fuels are often burned in the same manner. So, here is a typical burner and the fuel is carried through an inner pipe and air is carried through an annular pipe. Sometimes, this air may also be swell to enhance rate of mixing here. The air entrains inside the burning zone and a flame is formed. So, this is a typical laminar diffusion flame where the velocities are low. But if the velocities were high, you will get what is called as a turbulent diffusion flame with very jagged edges unlike the laminar flame which has a nice smooth edge. So, gaseous hydrocarbon fuels fuel flows through the inner pipe of the burner where the air flows through a concentric pipe. This is the situation at hand and essentially you have a free shear flow. There is no wall present across the thickness of the flow at all. So, what is the main objective? Now, here I am defining the fuel which is coming in through a pipe of diameter D. The jet spreads along the dotted line if you like whereas, the flame radius varies with x. It is highest at the beginning and goes on decreasing. When the radius goes to 0, you essentially say well that is the flame length L f. We are assuming stagnant surroundings completely stagnant surroundings and this is essentially the burning zone of the jet. The temperature profiles across any cross section would look like this. So, here is the radius and you can see that the value of the velocity there is 0 at the edge of the somewhere of course, the exact location where it will be 0 is not known and that is what we wish to find out. But, at the edge of the flame the velocity would be 0, but in the center it would be high. So, that is the velocity profile. The oxygen profile would go like that. The temperature profile would go something like that and the fuel profile would be as shown here. It will be highest and the edge of the flame is shown somewhere here. So, the fuel would be high at the axis. The oxygen would fly high at the edge of the layer and the temperature would be low in the environment, but would increase to a peak value somewhere and decrease a little in the core region. So, the main reaction actually takes place at the edge and therefore, you get very high temperatures there. So, the main objective is to predict the flame length L F and flame shape which means function R F flame as a function of X. In order to make life simple, we are again use the simple chemical reaction as the combustion model. So, in stagnant surroundings and assuming simple chemical reaction, I can write 1 kilograms of fuel plus R S T kilograms of air, oxidant air. R S T would be the air fuel ratio gives me 1 plus R S T kg of product. The postulated chemical mechanism is simply simple chemical reaction mechanism and our interest is to predict L F and R F as a function of X. So, now because this is an axisymmetric case, it is a round jet and therefore, this is an axisymmetric case. So, the continuity equation without constant properties would look as d by d x equal to rho m u r plus d by d r rho m v r equal to 0. Of course, in a free jet like that the pressure gradient term is 0. So, therefore, in the momentum equations you do not see any pressure gradient term. So, you are simply convection term and a diffusion term according to the boundary layer approximation. This is the equation for the fuel, this is the equation for the oxidant and this is the equation for the energy or the enthalpy H m and I would source term as r times r f u multiplied by heat of combustion. So, H m would now be simply represent the sensible enthalpy H m equal to C p m t minus t r f that is what H m would represent. So, these are the equations to be solved for this problem. So, first of all let us solve the velocity problem and we are going to make the very drastic assumption. We are going to say that the properties of the fuel or the properties of the mixture inside the flame zone is constant. So, properties are uniform. Then you will see that all these rho m's will come out and so would mu would come out here. So, I am now taking laminar flow. So, therefore, you have nu m by r d by d r r d u by d r is the diffusion term this is the convection term. The continuity equation would look like this. Now, again we can solve these two equations by similarity method where we define stream function x eta psi x eta equal to nu multiplied by f eta and eta equal to C into r by x and then u equal to 1 over r d psi by d r would become C square nu m by x f dash by eta v equal to minus 1 over r d psi by d x C nu m by x f dash minus f by eta. Now, the boundary conditions are since at the axis there is no v is equal to 0 you have f 0 also f dash 0 is 0 and f dash infinity is equal to 0. So, these are the boundary conditions for the similarity variable f. The substitution gives similarity equation. So, if we make substitution for u and calculate d psi by d y and so on so forth if we make these substitutions here. Then the two equations give us the following transformations in terms of similarity variable f f dash divided by eta square minus f f double dash by eta minus f dash square by eta equal to d by d eta which can be represented as d by d eta equal to d by d eta f double prime minus f by eta and combining these two can also be written as d by d eta of f double prime minus f by eta plus f f dash by eta equal to 0. So, if I integrate this from 0 to eta that is going from axis of the jet to some radius and noting the boundary conditions are f 0 and f dash 0 are 0. We get simply this will transform to f f dash equal to f dash minus eta f double prime and the solution then is f equal to eta square over 1 plus eta square by 4 f dash which is of interest. This is the velocity in the jet f dash equal to 2 eta over 1 eta plus eta square by 4 and f double prime is so on so forth and therefore, we can interpret u which is which can requires f dash and v which requires f dash and f in the following manner. So, u would be c square nu m divided by x into 2 over 1 over eta square by 4 square and v would be given by this expression eta minus eta cube by 4 1 plus eta square by 4 square c nu m by x. So, our next task is of course, to estimate what is c to do that if you multiply momentum equation by r and integrate from r equal to 0 to r equal to infinity then you will see that the momentum equation that we have this momentum equation you multiply by r and first of all write it in conservative form and then integrate from 0 to r then you will get d by dx of 0 to infinity rho m u square r dr equal to rho m v u r infinity minus rho m v u r 0 and likewise the gradient terms. Now, you will see these terms are absolute 0 because at r equal to infinity u is 0. So, that is 0 at r equal to 0 v is 0. So, that is 0 at r equal to infinity du dr is 0 because u itself is 0 and at the axis symmetry du dr will be 0. So, both this term as well as this term vanish and as a result all we get is 0 to infinity rho m u square r dr should be a constant. If I now multiply this by 2 pi then you will see this is nothing but the jet momentum rho m u square r dr into 2 pi r and integration. So, substituting for u which is this expression in here I can show that the integration would give me 16 by 3 pi rho m rho nu m square c square and that would equal since this is constant with x it will amount to rho naught u naught pi by 4 d square which is rho naught and u naught are the density and velocity at the entrance to the jet. This is a constant and therefore, I can now determine c in terms of jet momentum or I can also determine it in terms of Reynolds number. So, c is equal to root 3 by 8 Reynolds rho naught by rho m raise to 0.5 or in terms of jet momentum where Reynolds number is u naught d divided by nu m eta likewise remember what was eta eta is c into r by x. Therefore, if I substitute for c I would get eta equal to that definition u star which is and I can write u also as u star equal to u x by nu m in this manner in terms of Reynolds number and u over u naught would be 3 by 32 d by x r e 1 plus eta square by 4 raise to minus 2 rho naught divided by rho m. This is u divided by u at inlet it will be function of x and it will function of r and you can see as x increases u is decreasing at the same time as y as radius increases u decreases it will be maximum when eta is equal to 0. So, u over u max would simply be 1 over 1 plus eta square by 4 whole square and since we do not know where the edge of the jet will be we it is customary to define what is called half jet width. So, where u over u max will be equal to half eta would assume a value of 1.287. So, you can see 1.287 square divided by 4 plus 1 whole square would give you 1 by 2 and therefore, eta half equal to 1.287 is dimensionless jet half width. So, this is by convention of the dimension we say eta half characterizes the jet width r half. So, r half by x will be eta half divided by c equal to 1.287 8 by 3 Reynolds rho naught by rho m raise to minus 0.5 equal to 5.945 divided by Reynolds rho naught by rho m raise to minus 0.5 and that is nothing but tan alpha which is which what we call the jet spread angle you will see you will recall that I have shown you the jet spread jet spread. So, r half divided by x corresponding x will simply give you the angle of the jet it is obvious that if the bigger the Reynolds number smaller would be the jet spread angle smaller would be the jet spread angle as we would expect. Now, we want turn to prediction of length n r f. So, assuming Lewis number equal to 1 and making simple chemical reaction as we have done all equations can be rendered in conservative form of this type. We have done this many times and phi will be simply omega f u minus omega ox divided by r s t equal to h m in plus delta h c omega f u and it would also be equal to u or u naught because our momentum equation itself is of the conserved property form and because it does not have a pressure gradient. So, phi can now represents both the mass fractions it can represent enthalpy and it can represent velocity. Now, to locate the flame we define what is called as a conserved scalar phi equal to f. f is defined as phi minus phi a phi f minus phi a equal to omega f u minus omega f. Now, suffix a here implies in the air stream and suffix f implies in the fuel stream. So, omega f u minus omega ox r s t divided by omega f u minus omega ox r s t in the air stream same quantity in the f stream the fuel stream and the air stream. Now, you can see omega f u in the air stream would be 0 whereas, omega ox in the air stream would be 1 because omega ox represents air so that would be 1. So, this will be simply 1 divided by r s t omega fuel in the fuel stream is 1 whereas, omega ox in the fuel stream is 0. So, this will be simply 1 as you can see here and this would again be 0 and this would be 1. So, you get 1 by r s t. So, essentially mixture fraction f f is all often called the mixture fraction and can be given as omega f u minus omega ox divided by 1 plus divided by r s t plus 1 over r s t divided by 1 plus 1 over r s t. This is by taking phi equal to omega f u minus omega ox r s t f is given by that. Now, the flame is located where omega f u minus omega ox divided by r s t is 0 that is where the fuel and oxygen are in stoichiometric proportion and the stoichiometric proportion as we have seen means omega f u will be equal to omega ox by r s t. So, f equal to f stoic will be equal to 1 over 1 plus r s t that is the edge of the flame f will equal f stoic plus omega f u divided by 1 plus 1 by r s t inside the flame and f will equal f stoic minus omega ox divided by r s t divided by 1 plus 1 plus r s t this is follows from this. The flame edge would correspond to simply omega f u minus omega ox equal to 0 that would be equal to 1 over 1 plus r s t this shows the outside of the flame where f lies between 0 and f stoic whereas, inside the flame f lies between 0 and 1. To see this on graphically we draw a graph of f equal to 0 to f equal to 1 and here is what I have shown f stoic because r s t for a given fuel is known ox r to fuel ratio is known and therefore, f stoic can always be plotted on this f of course, being dimensionless fraction can only go from 0 to 1 and you will see that on the outside of the flame this is the variation of oxygen up to f stoic at the flame front the oxygen disappears. So, outside the flame there is oxygen, but it will disappear at f stoic the fuel is in the inside would go on decreasing from center line to the edge of the flame the product would go on increasing at up to the flame edge and then would decrease the temperature would be t infinity in the outside will increase to t at the f stoic and then would again decrease. Now of course, omega f u minus omega ox r s t need not necessarily be a single value there can be a range of values of omega f u and omega ox where the difference omega f u minus omega ox there by r s t is equal to 0. In fact, that is what is often found and that is what I have shown here that the fuel fraction would also appear little bit on the outside very close there to f stoic and oxygen would appear little bit on the inside and this zone is the flame thickness zone if you like or the flame edge thickness which you often see, but for all practical purposes we are going to say that the flame edge is a very sharp surface f equal to f stoic is a very sharp surface. Now, if we take phi equal to h star equal to h m delta h c omega f u where f is equal to h star then you can you will see I will get h m minus f delta h c omega f u the same quantity in the air stream and the same quantity in the f stream and same quantity in the air stream. So, this would simply be c p m t minus t infinity delta h c omega f u because remember there is no f u in the air stream c p m t naught minus t infinity again because in the air stream the temperature is t naught and in the fuel stream f u is equal to 1. So, you get delta h c here where there is no f u in the air stream. So, delta h c will be 0. So, h m in the fuel stream minus h m in the air stream is c p m t naught minus t infinity. So, thus noting that r f corresponds to eta f c r f by x and f stoic and r f is equal to 0. So, at x equal to l f we can say that phi is equal to u over u naught f h star is equal to 3 by 32 which is the solution that we had written earlier here. The solution to phi would be simply equal to u over u naught for all variables phi and that is what I have written here phi equal to u over u naught and all that sort of thing here. So, at r f if I replace eta by as eta f then I can recover r f by x equal to 16 by 3 raise to 0.5 r e into this density into that expression where f stoic is of course, 1 over 1 plus r s t and setting r f equal to 0. If I set that to 0 which essentially means this quantity is equal to 1 then x equal to l f I will get length of the flame divided by diameter equal to 3 by 32 rho m naught by rho m r e f stoic and f stoic being 1 over 1 plus r s t I get that. So, what this shows is that l f would increase with Reynolds number of the jet. The higher the velocity of the jet the longer will be the flame. Now, turning to turbulent jet flame let us look at what happens to flame length with increasing flame velocity and experimentally it was documented for the first time by a scientist called Hothon and he showed that in the laminar flames the flame length increases almost linearly with nozzle velocity as we have shown on the on the previous slide. We have shown here l f would almost linearly increase with velocity u naught, but after a certain velocity the flame length actually decreases and when the flow is completely turbulent the flame actually is independent of the nozzle velocity. So, in turbulent flows turbulent jet flames l f is almost constant the length of the flame is almost constant. So, we have something to think about here how we can predict the turbulent jet flames. The radial distribution of u is nearly uniform over greater part of the length and also experimentally it is very difficult to identify edge of the flame length because the turbulent flame is never steady it is unsteady and the edges are very jagged. So, you you get the flame which is sort of oscillating in the actual direction and it is very you can only take photographs of that to to measure what sort of time average flame length if you like as observed from photographs. So, the previous equations that we had used for laminar jet flame apply even for turbulent only thing is we have to use effective values of viscosity or kinematic viscosity and thermal diffusivity and mass diffusivity. So, we we say the equations of the three as I show here the equations of the slide 3 apply only thing is the effective values will be used as mu effective divided by Schmidt number and effective thermal diffusivity will be mu effective divided by Prandtl number and in gases Schmidt number and Prandtl number under turbulent range is point about 0.9 quite we have seen that in turbulence modeling part of the course. So, the simplest velocity formula is so we need now to define mu effective essentially. Now, the simplest formula for mu effective is 0.01 into rho m into u infinity minus u axial u axial is the velocity of the jet at the axis and u infinity is a co-flowing jet velocity. So, for example, I may have a jet which is like that and there is a parallel stream at u infinity sometimes you do get the co-flowing streams also. So, we define u axis here u x and u infinity then the turbulent viscosity would be given as 0.01 rho m u infinity minus u x delta this is the simplest form of the turbulent viscosity specification. It was proposed by Spaulding in the book combustion and mass transfer Pergamon press at Oxford 1979 and for stagnant surroundings of course, u x u infinity is 0 u x will be u max and also for from experiments it is found that the jet with delta thereby r half the jet half width is about 2.5 in turbulent boundary layers in turbulent jets and therefore, if I substitute delta for delta and set u infinity equals 0 then you will see I get mu effective equal to 0.0256 rho m u max r half and this as you can see r half can only be a function of x u max likewise can be a function of x and therefore, this is not a function of r. So, essentially what we have said is that mu effective is constant across the width of the jet it may vary with x we will see whether it does or not. From all our solutions we will apply if we assume constant properties and since mu effective is also nearly constant we will say simply change mu to mu effective and you will get u star max equal to u max by mu m equal to now r e turbulent square r half by x will be again given by that where r e is replaced by r e turbulent and what is r e turbulent it will be rho m u naught into d divided by mu effective which is 0.0256 rho m into u max r half. And therefore, from this I can get u max over u naught will be equal to 6.57 remember r half will get cancelled and 6.57 d by x rho naught by rho m raise to 0.5. Now, combining this with u star max I will get u max over u naught square equal to that or r half by x equal to 3.662 6.57 square is equal to 0.0848. So, r half is definitely a function of x in fact it increases linearly with x as r half divided by x 0.0848. So, in our effective viscosity formula r half will vary with x now let us see whether how u max will vary. Now, the results agree very well this particular result agrees very well with experimental data when x by d is greater than 6.5 because the initial range from just outside the mouth of the jet the r half is largely governed by ellipticity whereas, we have used parabolic assumption now replacing r half and u max we would have mu effective equal to 0.0256 remember I am replacing u max in terms of u naught d by x. In fact, you can see u max is inversely proportional to x and we just showed that r half is directly proportional to x. And therefore, mu effective is not at all function of x in fact it is absolute constant absolute constant. And that is why we our replacement of mu effective is perfectly varied all we are saying is a turbulent jet is simply a laminar jet with a much more augmented viscosity which is constant mu effective would then become 0.0142 I simply replace here u max equal to 6.47 u naught and so on. So, for an r half equal to 0.0848 x. So, x and x get cancelled and we get u mu infinity equal to an absolute constant as 0.01426 u naught d multiplied by rho naught rho m raised to half. Now also since eta half is 1.287 we can say that eta will be 1.287 r over r half and u over u max will be 1.414 r over r half whole square raised to minus 2. So, this is the velocity profile of a turbulent jet u over u max equal to all that and eta would be 1.287 r over r half and r half is simply 0.0848 into x. A turbulent flame now we have turned to L F and R F prediction a turbulent flame is essentially an unsteady and its edges are jagged as I said. Fragments of gas intermittently detached from the main body of the flame and flare outside diminishing in size turbulence affects not only L F, but also the entire reaction zone near the age of the flame compared with the laminar flame this zone is much thicker. We had identified that in laminar flame also there is a slight overlap where near F equal to F strike, but in turbulent boundary layers turbulent jet this zone is little bit thicker the flame so called flame zone or the age zone is somewhat thicker. This implies that if the time average values of omega f u and omega ox are plotted with radius r then the two profiles show considerable overlap around the crossover point F equal to F strike. Unlike the overlap in laminar flame which is caused by finite chemical kinetic rates however in turbulent diffusion flame the overlap is caused by turbulence and in the presence of turbulence R F u actually experience is not as high as estimated from R F u proportional to omega f u raise to x, omega x raise to y and this is because the fuel and oxidant at a point are present at different times. So, although the average values of omega f u and omega ox may be high the actual reaction rates R F u found is somewhat less and therefore how less is dependent on what is the probability of omega f u and omega ox meeting each other in the right proportions to cause a chemical reaction. So, the effective rates of burning are actually smaller than would be calculated from omega f u raise to x and omega f u raise to y which is the typical formula we use for burning rate of a fuel and therefore we must allow for probability of turbulence. Now, since all laminar solutions are applicable to time average quantities we may write phi bar equal to u bar over u naught equal to F bar over equal to h star equal to all that this is the solution repeated, but now let us see what actually happens. Let us consider the mixture fraction F and let this be the time average value. In reality the mixture fraction F would fluctuate between F high with a cap on top and F low with a cap on top these are the instantaneous values. So, what we are saying is that at the age of the flame in the presence of turbulence or because of the turbulent the mixture fraction would jump from a low value to a high value and high value to a low value almost instantly, but when it reaches a high value it will spend some time at in the high value and when it is suddenly fluctuates to a low value and again spends a little time there this is called the square wave or the rectangular wave if you like of variation. So, this is an assume variation of F. So, F is equal to F mean plus F dash on the positive side and F dash on the negative side. So, F dash can be both positive and negative around F bar and this is what I have shown here. So, although the F stoic of the fuel which is 1 over 1 plus r s t is reside somewhere here on the F and F equal to 0 and F equal to 5. So, the instantaneous high value may be here the average value will be in between the two, but that may not equal F stoic the average value of F may not equal the F stoic value. Now, what are the deductions we can draw therefore, for omega ox average omega fuel average and t average with reference to the figure with reference to the figure suppose that the value of F instantaneous F truly fluctuates between a low value F and I have value F. Let us assume that the fluid spends equal time at the two extremes and sharply moves from one extreme to the other. So, then F bar the average time average value F will be half of high and low values instantaneous values and F dash the fluctuation will be half of difference between high and low values. So, thus omega F u bar would be 0.5 omega instantaneous value of low and high mass fractions of fuel shown by the field circle. You can see this is this is the value shown by the field circle here would be greater than omega F u which is the value corresponding to the local value, whereas the time average value is over there shown by the field value and it is greater corresponding to F bar greater than F stoic. Now, we are taking the case of F bar greater than F stoic, but the story can be repeated even when F bar is less than F stoic. Likewise, omega ox is also greater than omega ox which is 0 for bar greater than F stoic. You will see that oxygen has already been consumed. So, the local value is 0, but the time average value is somewhere there omega ox time average value there. Therefore, for F bar greater than F stoic time average is greater than omega ox T bar 0.5 T instantaneous low plus T instantaneous high is less than T. You can see T bar is less than T the local value of T corresponding to F bar at the time average F. The above observations reveal will also apply when F bar is less than F stoic. Thus, in general finite amounts of fuel and oxygen are found when F bar is equal to F stoic. When F bar is equal to F stoic, you will get finite amounts of omega F u and omega ox at F equal to X stoic. Therefore, we get a little overlap. So, if F stoic does not lie between F l instantaneous and F h instantaneous, then T bar omega bar ox and omega will of course, correspond to F bar values. Therefore, the flame zone will be a finite volume enclosed by F l equal to F bar minus F dash when that quantity is equal to F stoic, which is the inner edge of the flame, flame edge rather and F high equal to F bar plus F dash equal to F stoic would represent the outside of the flame. Thus, F bar equal to F stoic surface will lie somewhere between the two surfaces. So, in this case, the thickness of the flame edge is now being analyzed. So, how do we estimate F dash? That is the issue. From our results of laminar flow results themselves, we can say that R f out minus R x will be F stoic minus F dash because that is the value of you can see for the outside F stoic will be equal to F dash plus F bar plus F dash and therefore, F bar would be equal to F stoic minus F dash. For inner surface, R f in would be F stoic plus F dash into all this and for the stoichiometric case, when F bar will be equal to F stoic, you will get R f stoic given by that. Now, we have to determine F dash. Now, F dash is determined from a mixing length formula and Spaulding recommends that F dash be evaluated as mixing length l m into d F bar by d r under stoichiometric conditions. So, F bar solution is already known to you. This is the solution to F bar. So, you take a derivative of this with respect to r and l m. Now, l m for turbulent jet, round jet is simply l m is equal to 0.1875 r half. For a turbulent round jet, this is found to fit the experimental data quite well. There is no distance from the axis term here simply because there is no presence of the wall and therefore, the mixing length becomes 0.1875 r half is essentially a constant. So, if we substitute for r half, then we will get F dash would essentially become like that. At each x, we can predict F dash R f stoic by x whole square. Now, if I set R f in each case, if I set that equal to 0, that equal to 0 and that equal to 0, then I will get Lf in sorry Lf out Lf in and Lf stoic. That is what I have done here. So, Lf out thereby d would be 6.57 F stoic minus F dash rho not by rho m. Lf in would be 6.57 F stoic plus F dash and since this is subtracted, you will see Lf out will be longer. Lf in will be smaller and the stoichiometric case where the in between length would be 6.57 F stoic rho not by rho m equal to 6.57 1 plus R st rho not by rho m. Thus, if Lf stoic is regarded as the mean flame length, then knowing F stoic equal to 1 plus R st raise to minus 1, the flame length can be estimated for any fuel. Remember air fuel ratio would vary stoichiometric air fuel ratio would vary for the fuel under consideration and therefore, we can say that we can readily predict the Lf stoic. Although the above relationships, the most important thing is you can see this relationship does not show effect of Reynolds number at all as was observed by Hawthorne experimentally that in turbulent flames, the length of the flame remains constant and that is what has been shown. So, we have we have recovered in spite of a very simple analysis, we have recovered the most important result. Now, though the above relations are only approximate, they do embody the form of the experimentally determined empirical correlations and what do they look like? The experimentally determined correlations for Lf show that F will be function of D as you have seen here, it will be function of R st, it will be function of rho not by rho m and rho not by rho infinity in some cases, the experimental correlation and rho not is essentially the density of the fuel thereby density of the surrounding gas rho infinity. But rho not by rho infinity is of course, the density ratio which can be different for different types of fuel in diffusion flame. So, except for that ratio, basically we have not been able to identify rho m and rho infinity because we assume constant property the ratio. So, with this I conclude the lecture on flames and with this I also conclude the entire course on convective heat in mass transfer. In the first 20 lectures, I covered laminar flows both external flows and internal flows with and without suction and blowing in the presence of pressure gradients and wall temperature variations. We also considered laminar internal flows both in simple ducts as well as complex ducts of non-circular cross section and we were able to calculate numbers both in the presence of circumferentially varying boundary conditions or actually varying boundary conditions. Then we moved to the next 10 lectures where on turbulent flows the formal aspects as well as turbulence modeling and also phenomenology called arguments of flow near the wall which gave us the universal laws of velocity distribution and in that of temperature distribution close to a wall which enables us to calculate the Nusselt number and friction factor for a turbulent boundary layer as well as turbulent ducted flow. Then we moved to the convective mass transfer problems in which we first of all postulated the we first of all considered the full boundary layer flow equation and we said there are simplified forms which are good proxies for mass transfer problems can be derived and the Reynolds flow model was found to be a very good proxy for the full boundary layer flow model and using that model we solved several problems. But of course, diffusion mass transfer is as important as convective mass transfer and for that diffusion mass transfer is simply a special case of a boundary layer flow model which we modeled as the stiffened tube flow model and found that the results from there were giving us the logarithmic form of the connection between mass transfer rate and the spaulding number b. In between these two Reynolds flow model and stiffened flow model we also invoked the queer flow model which showed us why property variations at large mass transfer rates are required and the model was successful in showing us the trends of property correction that should be applied and in the last two lectures I considered natural convection heat and mass transfer and in the present lecture I considered the case of heat and mass transfer in a free jet that is the case of a jet diffusion flame. So I hope you enjoyed these lectures and I also hope that this will prompt you to take up a career in the field of convective heat and mass transfer. Thank you very much and all the very best wishes to you.