 So, we have been moving on to trying to do something with the chemical kinetics data that we have been used to so far by trying to combine it with the thermal information that it goes with that means there is a heat transfer that is associated with the chemical reactions that we need to worry about and so can we put them together it is possible to put them together in some simple frameworks because we do not want to get involved in the flow processes and the mixing processes. So, the first thing that we will try to assume is it possible for us to actually neglect mixing and typically what do you mean by saying neglect mixing is that maybe mixing is not there it is not important it is not a significant extent or it is there a whole lot and almost lot of things are actually getting instantaneously mixed which is actually closer to truth in reality if you are trying to adopt these kinds of models simplified models where we are only interested in combining chemical and thermal processes without bothering about flow and mixing then typically you are looking at situations where the mixing is almost instantaneous right so you do not have to worry about it okay we will try to qualify this as when we are exactly neglecting versus when it is exactly instantaneous as we go along particularly for the plug flow reactor where we now could be looking at a simple flow description for the plug flow but for most other parts we should not have to worry about the flow because it is too complicated let us not worry about it either right so it is not as if like you do not really have a flow but when the when there is a flow it is too complicated for us to worry about maybe maybe it is not important for us to worry about that level of complication is it possible for us to deal without having to take into account flow description as much as we can and also by neglecting mixing if you try to adopt this kind of a framework where we are disregarding flow and mixing then we have roughly broadly two kinds of approaches one is what is called as fixed mass systems or open flow systems so the fixed mass reactors our goal is we now have a fixed mass of material what started out to be reactants in the beginning for the same mass of reactants the reactants now proceed to become products right so in a chemical reaction the mass does not change so it is the same mass that we are talking about but originally we started out having reactants and later on we are now having products when we first started adopting chemical thermodynamics right when we are looking at the equilibrium thermodynamics of the chemical reactions we started by talking about initial reactants and final products but we never really worried about the time evolution between the two okay for all we cared this could have taken ages in reality combustion reactions take take a jiffy it does not take a lot of time okay but whatever it is we would like to see how this evolution happens between initial reactants to final products through maybe intermediates right if you are able to take into account the detail chemistry as much detail as we would like to do right so this is what we are trying to do now as opposed to what we were trying to do earlier on with our equilibrium thermodynamics approach okay so in the in this sense this is not an equilibrium situation at all we are looking at a set of reactants that might be in equilibrium initially but you now look at departure from equilibrium as the reaction commences and then we now get into a situation where you finally get products which could be in equilibrium ultimately right previously we looked at only these two end states but now we are looking through these states so this is obviously going to be a problem that describes changes as a as an evolution in time therefore we need to have a time description of it what we will actually be looking for is ordinary differential equations in time for typically temperature and concentrations of of all the species okay so these are the two things that we should be looking for on the other hand in the open flow reactors we could afford to make an assumption of steady state which is more convenient for us we do not have to worry about evolution in time and that means everything that is happening is happening at all times okay so if you now try to do that then your time derivatives in your ODEs will go away and if you had ODEs in time and you had to get rid of the derivatives you now are stuck with only algebraic equations you see and that is actually like a direct consequence of applying this kind of formulation without the time derivatives you now get to get to algebraic equations which is the model called the well stirred reactor or WSR and but but you could still also do some more things like for it may be you say well I know how to solve ODEs okay I have done this here we are not going to solve the ODEs but let us suppose that we formulate the problem in the ODEs and we suppose that we can solve the ODEs well okay if you could solve ODEs in time why can I solve ODEs in space so if we have to solve ODEs in space it is ordinary differential equation in space that means I can allow for only one spatial variable right that simply means that I can now begin to account for one-dimensional variation in space okay so I could adopt like a one-dimensional approach which will lead to the plug flow reactor so you will now have ODEs in space so you will now have a D by DX kind of derivatives showing up over here in this so to picture is what we are talking about in the case of the fixed mass reactors you could do this as a either a constant pressure reactor or a constant volume reactor ideally this is this is idealization as you can see this is all these are idealizations right so or these are idealizations that are effectively simplifications so in a simplified system we now think about like a constant pressure fixed mass reactor so you can think about a template geometry which is like a piston cylinder arrangement where the piston has a constant weight in a what do you call spatially invariant gravitational field so we have to say that therefore we now say that this is going to be exerting a constant pressure the pressure is constant but the volume can change so as the temperature changes with whatever heat that you are going to try to remove out of the chemical reactions with the volume can change so this is this is volume as a function of time there so here our interest as I said is we are now trying to write ODEs in temperature and the concentrations of the species so we are looking for time evolution of temperature and time evolution of concentrations of the species okay so if you now started out with certain species like CI0 for I going from 1 to N and of course we will obviously start with reactants so if you want to number your species is 1 is equal to the first reactant this 2 is equal to the second reactant and so on 1 2 and so on will have nonzero values but if you now have your N N-1 and all those things is like products they would typically have 0 values right and typically you know even intermediates will have to start with you will have 0 okay so you will have nonzero values for reactants and zeros for intermediates and products to start with for CI0 and you also can indicate an initial temperature for this and then you now have the clock ticking T greater than 0 and then now it integrate your ODE system in time to get how this evolution happens as a offshoot of this you could now also try to find out how the volume changes in time all right and we will go through how to do that you have a constant volume fixed mass reactor where you know just to have a box okay you do not even have a piston that could move back and forth to maintain the constant pressure so the volume is fixed so the pressure obviously is going to change in time so that the problem is posed exactly similarly to what we have done before as far as the ODE set is concerned except we will now see some changes in the equations because you are now keeping the volume fixed rather than letting the pressure fixed and consequently as an offshoot we should be able to also find out how the pressure changes in time so this is how this problem would be posed on the other hand if you now look at the open flow reactors the first of which is the well stirred reactor it is in a way is easy a counterpart of this that means okay you could say you could say it is a counterpart of either of these as a matter of fact let us not worry about that but there is a lot of flow that is going on inside okay which we do not want to worry about that means we will not worry about the spatial variation of things inside how they are actually spatially distributed okay then we are also making the steady state assumption that means we want to actually look at how the system works under steady state conditions right that means we are not looking at a time variation as well okay so what it simply means is this is a model that is going to tell you if I put in reactance of certain concentrations here at a certain temperature okay with mass fluxes for each of those like this summed over to get you a certain m dot which is constant okay so the mixture m dot is going to be constant throughout the system right from inlet to outlet okay but the individual species m dots could change such that the mixture m dot is constant so sigma mi in over all I will be equal to sigma mi out over all I equal to m dot okay that is the way it this is going to work out the question is if I were to put in these reactants in at this particular temperature right what is it that I am going to get out for a given volume of this that is not a bad problem to think about right so if I have a box and then I am going to put in certain reactants it is okay for me to expect what to get out of this box right so we will find that this is actually pretty useful yeah and I will explain to you how useful it is in different contexts then as I said you could also allow for some simple flow description and the simple flow description is essentially in a one-dimensional sense what we will now allow for is a quasi one-dimensional variation case a quasi means that you will allow for like a area variation to happen along the flow direction largely speaking so your a is equal to a of x that is given right and you now allow for everything to vary as a function of x that means your temperature concentrations density pressure velocity okay because your area is changing the velocity will change below because the velocity changes the pressure changes the pressure changes so the density changes the density changes the temperature changes everything changes because of this all as a function of x nothing in a nothing as a function of time okay so we will start with reactants as usual it is a steady state situation so m dot is the same we measure things along x direction as a variation there so this is typically how we are going to deal with these different situations alright so let us now look at how to deal with yeah yeah which one is similar to that the combustor we use in a craft I will take up this question after I finish everything okay the answer is the combustor in an aircraft for example is a is a is a is a combination of about two or three wsr's and pfsr's put together in a network okay or in other words you could model your combustor that way if you do not want to get into actual physical details of how things are distributed and so on we could we could come up with a network in which you now say the primary zone is one wsr the secondary zone is another wsr and the tertiary zone like the dilution zone could be a psr or a wsr depending upon whether you want the spatial variation or not okay so reality is more complicated than these but these things could now be used as building blocks to create like a network that would simulate the reality alright you could think about this as something that can that can work easily for like a rocket nozzle for example so this is not a bad idea a well stirred reactor at many times works well for things like furnaces okay so wherever you have like a feed constant feed and then you have exhaust going out so you could sometimes or many times as a matter of fact model those kinds of complicated combustor geometries with a wsr liquid rockets yes sure that is possible okay we will look at why we need to have networks for like a like an aircraft aircraft combustor because we have a primary zone where you are having a stoichiometric combustion first and then you have like a fuel lean combustion next and then you now have a tertiary zone which is not too much of reactions and so on so that is the reason why you need to have a network otherwise you could actually have for a crude approximation you could have a combustor replaced by a wsr that is fine okay so our goal here is to find the evolution of T C I and V given initial conditions that is T not C I not V not so this is what is called as a initial value problem IVP so how would you do this essentially the broad outline for any of these approaches is as follows as far as species concentration is concerned we have to now apply conservation of mass of individual species okay as far as temperature is concerned we have to look at a global energy balance. So of course what will happen is you will find that each of these equations depends on all the variables that means equation for the ith species mass balance will depend on all concentrations concentrations of all species C I and T and the global energy balance will not only depend on T but also all the C is so you now get a system of equations right so this is what we will do and it is more straightforward for us to think about a global energy balance in terms of the first law of thermodynamics okay so you just have to apply the first law of thermodynamics right away so energy balance is first law of thermodynamics that means Q-Q dot-W dot equals MDU DT now this is treated is given right in a sense if you think about it if you were to if you are a partial differential equations expert and you were like raring to go on trying to get multi-dimensional spatial variation as well as temporal variation simultaneously all together right you will now say this is a box with these boundaries and then there are like boundary conditions that need to be given I need to tell what should be the heat flux here what should be the heat flux here on this wall what should be the heat flux on this wall and so on separately right but here we decided that we will not worry about any spatial variation right we are looking at only temporal variation so we are looking for ODEs in time and therefore this is like a some sort of like a collective boundary condition so it is like what is a net heat flux across all the surfaces okay so this is supposed to be like a given you can think of this as a collective BC you do not require a boundary condition at all here but if you were to have boundary conditions in a real problem how to convert that into a given property here is what you are looking for in this so I would say collective BC within codes now we have to look at what is W dot therefore it is not very difficult for us to fathom I mean W dot is essentially going to be like the expansion work right so let us look at how that comes about so enthalpy H equal to U plus PV right now most of the time in fluid mechanics a gas dynamics you would not really worry about what this H is okay but we are dealing with a multi-component system that means a mixture of species right and then in addition to that we also have chemical reactions going on so we will have to actually distinguish between standard heat of formation and sensible enthalpy for each of those species so this is pretty much now going to actually grow like a genie out of a bottle soon okay anytime you see a very innocuous H in your combustion class be prepared to see that it is going to now just grow like a genie right because it is now going to be taking care of all the species in there and for each of those species you have to look at the standard heat of formation and its sensible enthalpy so we will we will do that soon okay so long as we are not we are more innocent at the moment right we could simply write this as du over dt equal to dH over dt minus P dv over dt P is constant so we do not have to worry about changing it is changing with time so this is constant pressure constant pressure now then since since work is only in the form of expansion we can easily write that W dot divided by m that is the rate of work per unit mass of the system is equal to P times dv over dt right so from here you now put things together we already have du over dt is equals to dot minus W dot divided by m W dot divided by m is already P dv over dt and then we have du over dt is equal to dH over dt minus P dv over dt so putting putting everything together this simply means that q dot divided by m is equal to dH over dt this is something that we already seen earlier okay when did we see this the moment we actually define heat of reaction so when we said heat of reaction is the heat that was released in a chemical reaction which started out at a certain temperature and pressure and came back to the same temperature and pressure is what we said okay and at that time we noticed that the heat that is released during this process is the same as the enthalpy change okay and that is the reason why to find the adiabatic flame temperature at constant pressure we were equating enthalpies of the reactants and products that is the initial state to the final state and this exactly says that q dot is the heat that is released from the system okay and per unit mass H is actually per unit mass it is the reason why we have to divide by mass here okay small h is per unit mass I will explain this notation pretty soon it is going to get a little bit more complicated as I said right so the change q dot is the one that is making it equal to dH over dt if it were simply q that means you do not have to worry about the rate of heat that is going out it is a total heat over whatever time then you will simply have H okay so this is primarily because it is a constant pressure process that means in the constant pressure process the specific heat flow rate is equal to the rate of change of enthalpy okay so this this comes about mainly because of constant pressure I will contrast this with what you will get for q dot by m in the constant volume case intuitively we should expect that this should contribute only to the internal energy change because the pressure work is forbidden in a in a constant volume situation you do not have a you do not have room to expand right so we will be able to see the contrast in the first law of thermodynamics when you apply this to the constant volume case next but at the moment we will stick to the constant pressure case now H is now equal to H over m okay now this is the total specific enthalpy this is the total enthalpy right so when you say specific that means it is per unit mass okay the next thing I am going to say is something called molar specific okay when I do not say mass specific then it is implicitly per unit mass if I have to say per unit mole then I will have to specifically say molar specific okay so at the moment this is mass specific regardless of how you want to count your enthalpy okay the total enthalpy you want to now say mass times the specific enthalpy will give you the total enthalpy or number of moles times the molar enthalpy will also give you the total enthalpy the total enthalpy is the same right now so you are now going to bring in the molar enthalpy the molar specific enthalpy so that is given by sigma i equals 1 to n ni hi divided by n where hi is capital hi is the molar specific enthalpy of species I and ni of course is number of moles of species I number I am sorry number of moles species I or I at species whichever way you want to call it now see we use a capital hi for these molar specific enthalpy all right we will not really worry about a total molar specific enthalpy at all then so we do not have to worry about what symbol we will use for it okay so this is the notation here for what we are doing now you let us suppose we now differentiate this so differentiate differentiate this obviously with respect to time we are looking at what happens with respect to time so the mass is constant so you can happily take that out the denominator so this is constant that is what fixed mass systems are all about but unfortunately both the total both the number of moles of each species as well as the molar specific enthalpy of the species are changing in time okay why would they change why would well the number of moles changes because the reaction is happening okay so you start out with some so many number of moles of reactants at the end of the reactant you do not have them so that means that that has changed right and in the process some number of moles of products that did not exist before has been produced and so on so that is obvious why would the capital H I change temperature okay so capital H I the next thing is we are going to write this as the standard heat of formation plus the sensible enthalpy the sensible enthalpy depends on temperature so temperature is going to change in time as a reaction happens therefore the sensible enthalpy changes therefore the molar specific enthalpy is going to change so you have to take an account the change of both with respect to time so that is sigma I I am going to I am going to get tired of writing I sigma I equal to 1 to N every time so I am just simply going to the sigma I and then that is H I d Ni over dt plus sigma I Ni d H I over dt okay for ideal gas H I is a function of temperature only and that is the expression that we were talking about that is it is it has this standard heat of formation and the sensible enthalpy and the sensible enthalpy is the one that is changing with temperature okay. So if you are now looking at the derivative of this with respect to temperature it is a derivative of the sensible enthalpy with respect to temperature at a constant pressure okay what is that okay therefore dhi over dt is dhi over dt at constant pressure dt over dt right. So this is CPI dt where this is now molar specific heat heat of species I molar specific heat at constant pressure for species I that means if in case you do not think about this every species is going to have its own specific heat there is something that of course we noticed during the equilibrium calculations for adiabatic flame temperature okay just wanted to remind you second thing we are using a capital C for molar specific heat to go with the capital H for the molar specific enthalpy all right for the for the specific specific heat right that is per unit mass so this is this would be joules per mole kelvin all right that would be the SI units for this the standard one is joules per kg kelvin we will use a small c over there okay so so we have that and next so you are trying to evaluate these these terms here so next we have 1 over v d and I over dt equal to dci over dt equal to omega I this is how we defined these things if you remember right so you know run up to looking at the law of mass action this is how things were defined for you all right. So that is beginning to look at combining chemical kinetics and thermal information you see so this is coming from your kinetics say omega I is actually a function of CI, T you had that huge expression you see remember right that is like sigma I equals 1 to capital M for the number of reactions new I k double prime minus new I k single prime and blah blah blah some huge expression which had all the CI is thrown in there and the temperature in the Arrhenius expressions for each of those reactions right that is an ugly looking expression there we are spared of that because I am not going to write this out but it is there with you you know it is okay now therefore d and I over dt is equal to v omega I all right so plug these there plug these above above is here we can now say dh over dt is equal to q dot by M we can plug that for the dh over dt and for d and I over dt I am going to plug v omega I and for capital H I over dt I am going to plug CPI dt over dt right that is all I am going to do does not look like a big problem the M gets cancelled you have a q dot over M you have a 1 over M so the M gets cancelled so we get q dot equal to okay let us just write q dot over M equals 1 over M sigma hi v omega I plus sigma ni capital CPI dt over dt so in this we notice that of course the sigma is over I can I mean I am even dropping that okay but v does not depend on I okay that is that is common for all these species so I can pull this out of the summation similarly dt dt over dt does not depend on I so I can pull this out of the summation M gets cancelled so considering these I can go with q dot equal to v times sigma hi omega I plus dt over dt sigma ni CPI for this step q dot divided by v is equal to sigma hi omega I plus dt over dt sigma ni divided by v CPI what is ni divided by v number of moles of species I per volume is a concentration okay the molar concentration so this is q dot divided by v equals sigma hi omega I plus dt over dt sigma CI capital CPI so from here we can get dt over dt equals q dot divided by v minus sigma capital HI omega I divided by sigma CI capital CPI so we will call this equation one so what we have achieved here is to write dt over dt in terms of the concentrations and temperature is the temperature showing up here the temperature is showing up okay HI contains the temperature through the sensible enthalpy where HI equals HFI not plus integral T raft to T capital CPI dt okay and omega I equals omega I of CI small CI, T okay that is the same as what we have written there so we now have a fairly ugly ODE 1 ODE for temperature okay in terms of CI and T so now you look at how the CI is going to change right so you can get your volume so one of the things that is missing here is volume so we can get the volume volume V is obtained by mass conservation that is there is v equals m divided by sigma CI Wi where Wi is the molecular weight of species I and this is given okay fixed given okay you know what is the size of your system from your m from which you can estimate your volume which can be plugged in here I already told you Q dot is given and so you have an equation there see the evolution equations for composition for CI that is is obtained by consideration of what chemical reactions are going on so the CI is going to change when time so we should be able to get the change in CI because of chemical reaction what else what else is actually causing a change in CI CI is molar concentration and molar concentration means number of moles per unit volume okay so let us suppose that you did not have any chemical reactions but you had a bunch of species inside your piston cylinder arrangement with the concentrations change the pressure is fixed but something else the volume changes so if the volume changes will the concentration change yes as what concentration means okay if everything is concentrated in a small volume then the concentration is high that is what concentration means okay so many times we just simply forget our English right so the so evolution equations for CI is obtained by consideration of chemical reactions chemical reaction and change in volume right so DC over DT is Ni over V and this is minus Ni over V squared DV over DT now keep in mind we have already noticed that this is nothing but omega i all right so this is omega i therefore DC i over DT equal to omega i minus CI divided by V Ni over V is CI so one of the V's gets out there if you now write things in terms of CI DV over DT in the next thing we are going to do the constant volume case we can easily see that DV over DT will be 0 so this does not exist so you will simply have only this right so I am saying all these things so that you can we can ease that part now this is not easy because we know that omega i is a function of CI and T all right and then we have to look at what DV over DT is so consider equation of state PV equal to sigma Ni times or ut so this implies that 1 over V DV over DT equals 1 over sigma Ni sigma D Ni over DT plus 1 over T DT over DT right so if you now substitute here DC i over DT equal to omega i minus CI sigma omega i divided by sigma CI plus 1 over T DT over DT so this is your second equation now if you are unhappy with the DT over DT showing up in the expression for DC i over DT okay because it does not look like a nice set of ODE's you are free to plug in here plug this in there and make it look a little bit more ugly so long as you are happy with all the derivatives showing up only on the left hand side all right it does not make substitution of this there does not make them dependent on each other they are independent equations they are still independent equations all right so now have two equations and I should not say two equations I take that back this is n equations okay so this is i equals 1 to n, n is the number of species so number of species could be quite large for the hydrogen system hydrogen oxygen system you are looking at maybe thinking about like 6 to 9 species okay including intermediates of course but for something like a carbon hydrocarbon oxidation right if you are now looking at hydrocarbon oxidation by air that means air has nitrogen and then you want to allow for the nitrogen to participate in the reactions by forming oxides of nitrogen okay at elevated temperatures you are looking at maybe about 400 reactions and about 100 species okay so this is this is very very difficult equations okay so you know looking at about 101 equations to reckon with all right so happy solving yeah but we pose the problem yeah the problem is of the form dt over dt equals a function of ci, t dci over dt is another function of ci, t i equals 1 to n given t of t equal to 0 equals t0 and ci of t equal to 0 equals ci0 we also can obtain v of t from dv function of ci, t okay once you get your once you solve this you get your ci and t is a function of time you can get your volume evolution okay so this is how we will solve this problem.