 This summarizes what we have done so far on the well stirred reactor where basically we have considered a global mass balance over a control volume and global energy balance. So effectively now get a set of N equations which you can now say okay one equation is a sample for the ith species there which contains your yi out which could now be just simply called as yi as the N variables okay or the one sample variable for this equation but then we also understand that omega i is going to be a function of temperature therefore we need another equation to close this which is now the global energy balance and the global energy balance again involves the yi's okay so it is now coupled with this one so unless you solve this we can solve this unless we solve this we can solve that and then we now see that the hi's are the ones that are containing the temperature sitting on top here okay. So of course you could make a calorically perfect gas assumption and say this is like linearly in temperature and so on if you want to simplify your life but if you want to do the hardest work then you keep the integral do polynomial fifth degree sixth degree polynomials for Cp and go on integrate and all those things and so okay. So that is the two equations and two unknowns but however I would like to point out or maybe you should be asking this question to yourself omega i is not only a function of temperature but it is also a function of concentrations okay whereas concentrations of species okay. So this is actually a function of ci,t okay so but omega is a function of ci,t but we do not we have not really had ci's as our now unknowns we have been having yi's as our unknowns okay so we need to supply ci here right so there must be some sort of a relationship between yi and ci to close that set so you could look at this is you are now actually having ci is as another set of unknowns so it is actually 2n plus 1 unknowns and equations. So you need to have another n equations that relate yi and ci so that is basically given by yi is equal to ci w i divided by sigma j equals 1 to n cf c I am sorry cj wj okay. So this is like any time you try to find a yi you have to find a ci out of it and then plug it back into the omega i and use it there and so on so this is this is i equals 1 to n as well okay. So previously I mentioned that in some of the problems where we want to size the combustor right we want to know the volume rather than input the volume okay so the volume is now like the thing in question so therefore you now try to redo this problem a little bit by saying sometimes a residence time is defined as T R is equal to V divided by m dot how do you get this m dot times the time it takes is like the mass flow into it the total mass flow into it that should be equal to the density of the mixture times the volume the mixture occupies okay that will be the mass that is contained in it. So if you have a m dot kind of mass flow rate that is coming into this volume over a time of TR during which it is supposed to reside in this volume m dot times TR is the mass that is into this volume that should be equal to Ruby right. So with of course rule given by the equation of state where we now say this is equal to P times W mix divided by Ru T you now begin to see the molecular weight a lot more okay previously we were dealing with moles and we were always looking at the universal gas constant but now the molecular weight kind of tags along because we are now beginning to do things in a mass basis okay. So here W mix is the mixture molecular weight that is defined as it is not really defined you can actually calculate this this way so this is obtained by a mass weighted approach. So this is like 1 over sigma i equals 1 to n y i divided by W i this is again a statement of mass conservation okay. So if you take like one mole of the mixture and look at its mass constituents each of which has different molecular weights okay how will the number of moles of each of those pan out and then you now add the number of moles and so on okay. So that is how we are trying to do this this can be derived but I am not going to derive this for you just stated it. So if you do all these things then here in this problem we are given we are given m dot that is like the throughput that we want to achieve v or tr okay and q dot so if you now have tr you can try to find out v and so on of course I guess we also are given p so in NP so the pressure at which you want to do the work to do this to this problem and then you can actually have a relationship between so T is a variable it to be evaluated and so you can actually have a relationship between tr and v okay and what else that is about it okay. m dot is given this is given that is so you just have to go ahead and solve this okay. So we have so if you want to now count 1 let us say 2 and 3 okay 1, 2 and 3 actually gives you 2 n plus 1 equations we are interested in obtaining yi, T that is n plus 1 quantities right that means if it is possible for you to eliminate your C is okay for 2 n plus 1 equations for yi, ci and T that is 2 n plus 1 variables just as well but we are interested in these ci and yi they do not really give additional information okay but you just have to go through that. So what typically happens in this case is you could do this problem in multiple ways depending upon what you are looking for as I said the other day okay so essentially these are algebraic equations you notice that so 1, 2 and 3 are all algebraic equations but they are nasty algebraic equations okay. So this is going to be a very nasty looking expression and temperature first of all and then you have this T sitting on top of this integral and all that stuff so this is not going to be a very easy equation to a set of equations to solve okay. So what would we do with this we can do a couple of different things one for example you now say let us look at these parameters in the problem all right and if I am now allowed a certain m dot I fix that and certain p and I now say I will allow a certain q dot that is fixed okay that is like a heat loss to the surroundings if you want or if you want this reactor to be heating something else like let us say in an external combustion engine where you have a combustion chamber that is giving heat to like say let us say a water to get converted into steam or something like that you want certain q dot okay. So let us suppose that you can put that there and so on now but you want to make sure that your combustion is complete okay. So what you then do is you now start varying your TR okay so as you now keep on varying your TR you now get for each value of TR you now get your Y I and T and then you now can plot your Y I and T as a function of TR okay Y I and T are actually not functions of pretty much anything there are no independent variables here we are not looking at time or space nothing okay these are parameters in the problem. So you now vary your TR you can now get your species and temperature to be calculated as functions of TR and then plot this what you should find is beyond a certain value of TR these things will stop varying right you will also think about it like it is as if like giving more and more volume is it essentially it is like you now have like a let us say large pipe and then you are basically looking for how long this pipe should be for your reactants to become products larger the longer the pipe larger the volume and greater the residence time. So if the reactants are actually become product somewhere here but your pipe is still longer okay the products are going to continue to remain that way and what you are going to get out will be the same as if you were to add truncated your pipe somewhere here you see. So if you were to increase your TR your answers are not going to vary further on right that means you can actually decide what should be the right residence time by the time these things now level off to some value beyond which you do not have to worry about. So that is like the way you size your combustor you now say this corresponds to a certain volume and therefore I will now get my length for this diameter and so on okay we cannot distinguish between diameter and length that is a no no because we do not know how what is happening inside okay so you fix your diameter by some of the consideration or the lateral distance then you can get a length out of it and so on because you can estimate a volume for complete combustion okay. So one of the ways of doing this is by beyond a certain tr or v yi and t stop varying this means no further change no further change in the reactor for the extra volume or residence time provided right. So this is used to size the combustor or okay we will just stick to the word reactor as if you are chemical engineers okay that is fine another thing that you could do is let us suppose that I had a reactor all right let us suppose that I had a reactor and my question is how much m dot can I can this reactor handle can I dump more and more reactants at it all right so that is another way of looking at it so essentially the point basically these are parameters you could vary them and you can now fix a few things you can vary one of them to see what is the effect of it so another meaningful variation is to actually keep varying your m dot given other things okay so beyond a certain m dot I mentioned this earlier beyond a certain m dot the equations do not converge to a solution when you say converge it implicitly means that you are going to solve these things numerically okay with an iterative approach it is like this is like check and enact problem I do not know omega I I cannot get my Y I so I do not know my Y I so I cannot get my T and so on right and do not even think about solving this simultaneously like you used to do in your middle school simultaneous simple algebraic equations to equation to a known system right so obviously what you are going to do is say iterative we start with some guess right and then plug that in there and then see get an answer for the next iteration and then and so on and of course if you keep doing this blindly you are probably going to go haywire okay so your starting guess must be pretty smart okay and then you need to look at some things like method of steepest descent towards the solution and so on so you have to adopt even as a matter of fact Newton-Raphson method kind of thing is sometimes too simple for these kinds of solutions okay that there are required here so ultimately but basically what I am saying is you are looking for a numerical solution that means you have to converge to a set of answers for you so you may not converge all right so that that simply means as far as this model is concerned is a blowout that means these equations just simply cannot be satisfied for the set of parameters that we are looking at all right the typically the way you do this is you start with some m dot and then keep varying okay until you hit a point where you cannot get a solution then you know that beyond this point the flame won't sustain this is essentially what it really means okay so that becomes a meaningful exercise it is like it is not like you just take took some set of parameters here and then just juggle with varying them taking like two m dots and like let us say take two pressures and all those things that is not the way you do you have to systematically do so that you approach the no solution situation and then you know that you are approaching m dot in a physically meaningful manner right okay I would like to point out a couple of things in what we have done here which are which are somewhat unique or interesting one the notion of residence time is a very powerful idea okay regardless of how complicated an analysis that you want to do or not this gives you this gives you a first cut idea so for example one of the problems with scramjet combustors like supersonic combustion okay is lack of residence time or inadequate residence time so you are looking at a very fast flow that is coming in which is decelerated to let us say Mach 1.62 something like that 1.62 something of that sort I still you are looking at supersonic flow and then you now have to inject sometimes if you are unlucky you have to inject liquid fuel and then that has to actually get atomized and then get into droplets and then vaporize and then mix and then burn okay so keep doing all these things you are out of the combustor where the flow just does not want to stop alright or if you really stop the flow for all these things you are not going to fly okay so the residence time is one of the key ideas in most of the combustor designs that you want to think about okay. So this is a powerful idea that comes out of this analysis the second thing that I would like to point out before we get to the next thing which is the plug flow reactor is in whatever we have done so far in combining the thermal and chemical processes so far which is constant fixed mass reactors both constant pressure and constant volume and the well stirred reactor on an open flow system is we have primarily considered only two conservation laws one is energy conservation the second one is species mass conservation okay in some sense you will find this approach recurring as a recurring theme in most combustion problem approaches in other words you always think about doing the combustion part what is special about combustion in addition to just a flow situation is primarily two things one first of all you have to deal with multiple species and therefore you have to conserve the mass of each of those species. So you now are having to deal with a species conserve a set of species conservation equations in addition to that you it is directly coupled to the energy balance because the reaction rates depend on temperature and you need to know what the temperature is through the energy balance which in turn will depend on the species for the heat release so they get coupled okay many times you will find that it is possible to decouple this set of equations from the flow set of equations therefore then you will now be able to identify that the flow equations are like a standalone and the combustion equations can be solved after you have solved the flow equation and plugged in the velocities here for the convective effect of the species and the temperature and the enthalpy okay so in some sense the you can now say that is a flow problem and this is the combustion problem. If you were to basically look at something like convective heat transfer of incompressible flows you will understand that this kind of thing happens there and there what you would say is you can solve for the flow which is decoupled from the energy balance and you can now plug the flow field that has been solved from the flow into the energy equation to get the temperature field and from there you can get the heat transfer that is heat transfer kind of approach this is similar here except that in addition to the energy equation which is going to be complicated by the presence of multiple species and reactions and so on you also have to take into account the species mass conservation equation set and this constitutes the combustion problem mostly so most of the time we will always be worrying about two things species mass conservation equations and energy equation this is like pretty much the combustion problem there are some occasions where this will get tightly coupled with the flow problem there are lots of other occasions which we will be looking forward to where it gets decoupled therefore we do not worry about the flow problem we outsource that to a fluid mechanics person okay so to speak you get the flow field and then you can solve the combustion problem so what we have been doing is the combustion problem the reason why we I am stating that now is the moment you actually step into the plug flow reactor we are now going to actually keep it keep track of what is happening in space and what is going to happen in space well all will primarily be influenced by how the velocity of the flow is going to change in space in relation to how like let us say a quasi one-dimensional manner the deduct area changes okay so we will go back to something that is familiar in our gas dynamics where if you now have like a area decrease the velocity increases and so on so the velocities change so the velocity now comes into picture that means we have to bring in a momentum equation okay whereas we have never bothered about momentum so far okay so typically this is combustion where we do not have to worry about momentum equations but the plug flow reactor we will do that okay so let us now look at the plug flow reactor PFR so as I said if you wanted to have PFR rhyme with the well stirred reactor you would want to call it perfectly stirred reactor and call it PSR okay but this is PFR no no no changes in how we want to do this so here the model that we are looking at is you now have a variable cross sectional area situation so a equals a of x where x is going in the predominant flow direction we do not want to entertain any other flow direction the flow is actually turning around and all those things to go along with the walls and so on we are oblivious to it we would we will keep our eyes shut okay instead we will actually be looking at a velocity which is a quasi one-dimensional velocity the quasi one-dimensional velocity is not a real velocity okay it is not directly it is not the same as the axial component of the actual velocity because in reality when the flow turns there is a certain mass that is associated with going that way okay but we will conserve mass as if things are all going only this way okay so our quasi one-dimensional velocity will have to conserve mass only in one direction so that velocity is not the same as the axial component of the actual velocity okay so this is the way we will do this of course many times I find that gas dynamics teachers who actually deal with this do not emphasize this particular point and then we always get confused when we want to get back to our reality comparing experiments and so on okay we will get discrepancies the reason is we will have to keep this in mind okay so what we will do is again look at like a local control volume and look at something like a q.double prime notice the double prime that is coming up we never had a double prime so far okay so what this really means is this is the heat flux so far what we have had is on the q.is in is in vats okay this is basically the rate at which heat goes out this is duals per second there is vats this is like the power this is heat flux that means it is vats per meter squared so since it is meters squared you have double prime okay so directly goes with the number of linear dimensions over which you want to normalize okay so if you had a q.tipple prime you can imagine it is like a volumetric heat loss or heat release whatever okay that is typical of this this is the way we would not take the heat flux okay so here we are saying that you now have a velocity that we will have to worry about keep in mind we are now introducing a v this is what velocity okay previously we have been using a small v that stood for 1 over rho that was like specific volume we now change okay the same symbol is used as we very very very rarely do you come across a formulation that uses both the specific volume and the velocity when you are now getting into things like velocity you are looking at a open flow system and you will always use rho okay you do not have to worry about specific volume at all and things are all done on a mass basis you will not really use the equation of state on a molar basis and so on so we should not have any confusion from this point on so similarly we now say p equals p of x t equals t of x rho equals rho of x yi equals yi of x okay i equals 1 to n we have to keep track of spatial one dimensional spatial variation of n species long okay that is that is how we want to do this okay so assumptions first is to reflow what goes in should come out right away right away is the key point that means there is no cumulation or depletion okay quasi 1 b flow that means the v velocity is equal to v of x i vector so this v of x that we are talking about is a quasi one dimensional velocity it is not the axial component of velocity in the real flow we now assume no mixing in the axial direction all right that is to say yeah of course we are keeping track of things as a function of x right so if you now had a yi as a function of x and yi changed as a function of x then you will have yi gradients in yi okay so you will have gradients in concentration so if you have gradients in concentration does this mean that it is going to drive some mixing as we will see soon okay because mixing is driven by concentration gradients similarly temperature gradients will lead to conduction okay velocity gradients will lead to viscous effects we won't worry about that okay we won't worry about any axial transport processes in this formulation okay transport processes me meaning species mixing viscous effects and heat conduction okay so that is for mass momentum and energy transport we won't worry about any of those and since we are actually saying that the velocity is only in one direction we are also assuming that there is a instantaneous mixing that is going on in the perpendicular direction all the time that means we are oblivious to any variations of anything in the transverse directions let alone concentrations that means the concentration that you are looking at is supposed to be represented representative of the entire station it is supposedly the same along the entire station that means it is all got mixed okay no gradients okay so that means this actually assumes instantaneous mixing in the transverse direction right this is what I said right in the beginning when you are neglecting mixing it is because we say that it is not there or too much of it is there either way okay so you can you can neglect something if it is not there you are not neglecting there is nothing to neglect okay it is not there so you cannot even neglect okay essentially as far as the axial mixing is concerned you are saying it is almost negligible why would you say that that is because we are looking at a convective situation okay so even if you thinking that for example you had a gradient and therefore you will have a species that is trying to actually mix okay you have the convective effect that is bringing it back. So axial diffuse axial diffusion at axial mixing typically loses when you now have a reasonably large convection and this is given by Peckley number similar to how you would use Reynolds number to decide between inertial effect and viscous effects okay for the momentum balance okay so when you now have a significant amount of flow we do not have to worry about this that is to say it is small it can be neglected the other thing where you are saying that you are having an instantaneous mixing and therefore your mix you are neglecting is to basically assume that you have too much of it okay so you are you do not worry about it so this is also true in our lives when we have too much of something you take it for granted you do not even think that it is there okay so if you have like Wi-Fi everywhere you do not realize it until it should get shut off right or you have electricity all the time you do not realize it until it get it gets shut off so it is sort of like that okay so these many of these assumptions you know are very intuitive you just have to think about how it compares with your day-to-day experience and then you will understand that this is not this is not far fetched at all okay so that is what you that is what you are trying to do of course as I said the fourth assumption would be inviscid flow inviscid flow and fifth is perfect gas but I think by now you never questioned the perfect gas of assumption you have been brainwashed into supposing things are perfect as far as the gas is a concern yeah actually that is quite true if somebody were to ever question I mean is it really perfect most of the time yes the answer is do not worry about it okay most of the time in whatever we are dealing with the gases are indeed perfect this is one of the perfections that is relatively true in reality so this is actually a good assumption okay so as I said earlier we do mass conservation and energy conservation as part of them as part of the combustion problem but here we also have velocity okay so we have to worry about momentum conservation so mass conservation we simply have to say d over dx of rho u a equal to 0 quite familiar to most of us okay from some things like gas dynamics okay and notice that we are using ordinary derivatives okay it is as if like we do not even know how to write a partial derivative at this stage okay so we go step by step we are like with regard to learning combustion we are babies so we take baby steps towards partial derivatives just wait for a few more classes then you will find too many partial derivatives for you to stomach okay so let us be happy with the ordinary derivatives at the moment so x momentum conservation okay this is simply boiling down to your one-dimensional Euler rho u I should say I think I may I will just say okay this is fine this is fine we will call this that say that essentially says the velocity is only a function of x we will now call that as u okay so that is what you are using rho u du over dx equal to 0 energy conservation u a h plus u squared over 2 plus q q dot double prime times p this is capital P not the small p which is pressure so this is equal to 0 where capital P is the local parameter okay that is P equals P of x okay the parameter here refers to whatever happens here we are only keeping track of one dimension it could be like a box could be like a tube circular tube whatever it is so you have a parameter there that goes with it or in other words we are saying that the heat transfer is happening only along the walls we need to have an idea of how the parameter happens there the wall then we have the species conservation that is rho u so let us let us fix this up like this rho u du yi divided by sorry d yi divided by dx plus omega i wi equal to 0 this should have been like this together should have been small wi that is for the source okay the source essentially is like the number of the I am sorry the amount of mass per unit volume per unit time is what we are saying so this is mass flux times a mass fraction derivative alright so the mass flux times a mass fraction derivative will give you a volumetric rate of flow okay so this is essentially volumetric rate of flow omega i is the number of moles per unit time per unit volume times its molecular weight will give you the mass per unit time per unit volume okay notice something here we are not okay I have not derived this for you okay I am just writing this out and I am not going to worry about how to derive this okay we will derive a the general species conservation equation in 3d pretty soon okay taking into account diffusion and all those things but at the moment I just assume that you can figure out how to derive this let us not worry about it what I told you was well we were here I said typically we worry only about these two the energy and species okay and then I said because we are interested in the flow we have to worry about the momentum what is this and how is it different from this right this is mass conservation for an individual species so this is actually going as i equals 1 to n this is n equations anytime you look at a species conservation equation keep in mind it is actually n equations okay this is one equation for the mixture all right and that does not really have any production terms it does not worry about this kind of a term why is that because in chemical reactions mass is conserved so whatever is getting produced must be something that correspondingly gets consumed in the mixture all right therefore we should not have to worry about it we will come back to this much more specifically when we do the detailed three-dimensional species conservation and derive the mixture mass conservation from there okay but ultimately what would mean is the mixture does not even know that reactions are happening inside okay it is as if like for example when you are doing when we are doing aerodynamics let us say okay let us say you go to an aerodynamics class and then the aerodynamics professor is talking about flow past an airfoil and all that stuff and then he writes equations for mass conservation continued equation and so on you cannot say stop you are doing aerodynamics that means this air flow past an airfoil air contains oxygen nitrogen all these things these are all individual species why was the individual species conservation equations should really do that would you you would not have to the mixture does not know that it has it is actually a mixture even if nitrogen and oxygen were to react that is that is a point okay so it is very interesting there this is like this actually says something like a collective behavior okay as if the mixture than exists this is collective behavior with the mixed with taking cognizance of the mixture so it becomes it is part of the combustion problem so that there is a difference between taking cognizance of the mixture and not taking cognizance of the mixture so well so how can we do this how can we now deal with this so we can we can write we now have to actually do some more jugglery and I hope we can do this now pretty soon we could write some of these equations as let us say you want to pick the first one you open up these brackets there and so you write this is 1 over rho d rho over dx plus 1 over u du over dx plus 1 over a da over dx let me go a little further out and that right equal to 0 you now begin to get the drift of why I am having the 0 right at the far end I want to point out that rho and u are unknowns here but a is not an unknown it is given so 1 over a da over dx is like a source term okay it is it is known to you it is not an unknown okay then the next equation you can now write this as rho u du over dx plus okay I am just going to write dp over dx equal to 0 why am I spacing things out I am beginning to see if I can write all the like derivatives one below the other and I am going to now form a matrix okay so the next equation you could write this as u du over dx plus dh over dx equal to 0 right but for the other ones you have to do some jugglery so let us try to do that quickly this is lot of algebra so not much point in spending time so we notice that h is equal to sigma yi hi over you are using small catch use yi when you are using capital H you have to use ni ci on a molar basis so dh over dx equal to sigma dy I over dx times hi plus sigma yi dhi over dx this is equal to sigma d dy over dx times hi plus sigma yi partial du hi divided by partial T at constant P times dt over dx we want to bring in temperature okay so not contented with just enthalpy therefore we can write this as sigma dy I over dx hi plus sigma yi small CP I dt over dx the dt over dx can be pulled out of the summation and then you have a summation yi CP I so this together is basically CP for the mixture okay so this is equal to sigma hi dy I over dx plus CP dt over dx keep this then you have another problem P equals rho Ru T divided by W mix so this means that 1 over P dp over dx equal to 1 over rho d rho over dx plus 1 over T dt over dx minus 1 over W mix d W mix over dx oh my god can you now have to worry about how the molecular rate of the mixture changes with space okay terrible well of course we know how it changes it is supposed to change with the composition and the composition is given by yi so we should simply be able to write the variation in the molecular mixture molecular weight in terms of variation in the composition therefore but W mix is equal to sigma yi divided by W I hold the minus 1 so d W mix over dx can be written as minus W mix square sigma 1 over W I dy I over dx so therefore dp over dx is 1 over rho d rho over dx plus 1 over T dt over dx plus the negative signs get together so you get a plus W mix sigma 1 over W I dy I over dx all right therefore what happens is you can plug this dp over dx over here okay dp over dx but it is going to now contain d rho over dx dt over dx and dy I over dx right so and of course the species equation for you already had a dy I over dx and this acts as a source term okay involving yi's and T of course therefore it is coupled and we can write so we obtain equations in terms of in terms of well simply say rho u t and yi so these are like four unknowns but keep in mind yi is actually a set of n unknowns and so you need to have a four set four equations where the last one being a set of equations that is exactly what we have okay. So solve this and love happily ever after I will see you Monday.