 Now today look at the constant volume fixed mass reactor we looked at the constant pressure fixed mass reactor earlier will follow pretty much the same approach except that we now notice the volume is going to be fixed and instead the pressure is going to vary and correspondingly the analysis is going to change a little bit and we will have to see how that changes going to go through this a little bit faster when compared to what we needed earlier so that we just look at the modifications that we have done that will do now. So the first law of thermodynamics applied here would be to start with we have the same equation but right away we can now say this is equal to 0 because you do not have any pdv work pv work okay and nothing else okay so you just say there is no work done by the system or we just have only a heat interaction and obviously then any change in the internal energy of the system directly comes from the heat or vice versa okay or the heat that is released goes to changing the internal energy directly so here you now have du over dt equal to Q dot divided by M so all heat goes to changing the internal energy because it is constant volume and then we have a similar situation like what we had before that is we will now say we have a U this is a total specific internal energy this is now capital U divided by M this is now total internal energy right previously we had a small h is equal to capital H divided by M you are instead of that we are now looking at internal energy so this is sigma i equals 1 to n ni capital Ui divided by M now capital Ui is a molar specific molar specific internal energy of species I ni of course continues to be the number of moles of species I so if you are now looking for du over dt right so du over dt now changes because of two things one is the number of moles that changes with respect to time as well as the internal energy of species I that changes with respect to time so this is 1 over M sigma I capital Ui d ni over dt plus sigma over all I ni d capital Ui over dt there are two contributions here again we notice that the molar specific internal energy is a function of temperature alone so for for perfect gas perfect gases therefore du I over dt is equal to at constant volume here dt over dt now this is equal to capital CV I dt over dt so this is molar specific heat of species I at constant volume now if we can make a now plug du o du I over dt as CV I dt but then what about d ni over dt we can do the same thing is what we did before again 1 over v d ni over dt equal to omega I so this implies d ni over dt is equal to v times omega I therefore plugging all these back in the first law the means we say we recognize that du over dt is queued out by M we recognize that the du I over dt is equal to CV I dt and we recognize d ni over dt is equal to v omega I we plug all these three back in the original equation and you can now get q q dot divided by M equal to 1 over M sigma I capital u I omega I times v plus sigma I ni capital CV I d capital T over d small t notice here again as before that we does not have an index I so that is that is a common that is common for all the species and similarly dt over dt is independent of species they can be pulled out of their respective summations so and then M gets cancelled for nonzero M for those of you very picky about these so we get q equal to v times sigma capital U I omega I plus dt u over dt sigma over I ni capital CV I so from here we can rearrange to get finally I am going to skip a couple of steps which are just algebraic and get this as q dot over v minus sigma over I capital U I omega I divided by sigma over I ci capital CV I which I want you to understand is now a function of CV ci and T okay why because omega I depends on ci and temperature and you also have an explicit presence of Ci showing up here therefore this is a function of the concentrations and temperature of course you can we typically do not deal with specific molar internal energies we typically deal with specific molar enthalpies you look at these tables like Jan of tables and so on it typically less enthalpies rather than internal energies so you could now convert that so for ideal gas ideal gases ideal gases U I equal to hi minus Ru T and CV I equals CP I minus Ru okay so dt over dt is equal to q dot over v plus Ru T sigma I omega I minus sigma over I hi omega I divided by sigma I Ci capital CP I minus Ru the reason why we are actually having a universal gas constant here is because we are dealing with things on a molar basis okay so you would have a specific gas constant only if you are using things on a mass basis therefore this is appropriate I like to make a few comments we will we will in the future look at these kinds of expressions or terms but I want you not to get confused at that time sigma I omega I is nonzero primarily because the number of moles in a chemical reaction is not necessarily conserved it is a mass that gets conserved so if we were to actually look at omega I as the amount of mass that is produced per unit volume per unit time okay for which we will assign a symbol later on okay then sigma over all I of wi will be equal to 0 because you do not have any net mass that is produced or consumed if you now try to sum over all the chemical reactions sorry chemical species that are produced and consumed on their mass basis okay but that is not true when you are now trying to have omega keep this in mind because later on we will be looking for a sigma wi and plug it as 0 it should not get confused okay the English alphabet gives you 0 the Roman does not give you 0 so if you want to quickly remember that way that is also fine Roman is molar English is mass this you can now write hi as hf hf not I okay plus sensible enthalpy okay the hf not I times omega I that combination sigma I hf not I omega I will be the net heat release rate in the chemical reaction okay so it is finally coming from the standard heat of formation times the rate at which it is formed for every species added all together is the total heat that is released in the chemical reaction okay so and then the then you will have a sensible enthalpy part that that that will stay so these things will show up as we go along more explicitly we will just do not worry about it now just disturb so this as before is a function of Ci, t keep that in mind now here since the volume is constant 1 over v d ni over dt is equal to omega I can be directly written as dci over dt is equal to omega I previously we could not do this okay previously we had to notice that ni divided by v is equal to Ci and we will change with time so we had two terms for contributing to the rate of change of concentration okay one because of the chemical reactions the other because the pressure was constant the volume was changing and since the volume was changing the concentration was changing but since the volume here is constant we do not have to worry about that effect okay so the concentration changes only due to the chemical reactions here okay or in other words you could take the v directly within the derivative without any effect and get the concentration right there and that is what we are looking for we are looking for a time derivative of concentration as a ODE so this is again a function of Ci and t this is because of the chemical chemical kinetic equations that we saw earlier on so this is like a fairly large expression for each of those and n equations there okay and this is like typically what you would get solved in like with the package like Kimkin okay so you are you advance in time and see what happens but now what we said that is where the constant temperature but now what you are saying is we will now have this effect the temperature okay and then the temperature will change that is essentially the idea that you are talking about and so here DP over DT is of interest as a derived quantity that means we now have these two as a close set of equations we have n plus one equations and n plus one unknowns where n unknowns are Ci and n equations are these okay so this is the you should say really this i equals 1 to n and t is the n plus 1th unknown and this this first equation here is the equation corresponding to that but they are all coupled all of them are coupled so once you do this you should be able to find out DP over DT which is of interest here so the way we do this is to go back and say PV is equal to sigma over i ni or ut so we DP over DT is equal to or ut D by DT of sigma ni plus or you sigma ni DT over DT that means there is a contribution because of change in the number of moles and a contribution in because of change in temperature with respect to time towards rate of change of pressure so this gives you DP over DT equal to or ut take this V over here right and then get this derivative inside the summation also the volume inside the summation you will now have a sigma over i 1 over V D ni over DT which is nothing but omega i okay so this you this gives you a sigma over i omega i plus you can pull out the DT over DT from the summation get this volume over here and then that becomes a concentration so you get or you DT over DT sigma Ci right so that is a function of Ci and T which can be evaluated once you have solved for Ci and T this is not coupled with the other okay so you can get this so this again formulates the constant pressure sorry constant volume fixed mass reactor to solve this is not a joke okay if your n is a fairly large number should say fairly I should even say a fairly large even it is modestly large it is going to be complicated you got to pray to pray to your gods that n is like one okay then then you are in decent shape but even then it is it is it is difficult because this this term here that is showing up here and here is going to be so huge you are going to have the temperature sitting on top of a tree like an exponent to the e-e over RT there okay so that is very difficult to handle and there are numerical techniques for these okay so you have to use like some stiff reaction solvers so stick stiff equation solvers so there are there are special numerical techniques that can handle these simultaneous equations so let us now look at the other two ideas that we had on open flow systems so the first of those is the well stirred reactor which is typically abbreviated as WSR it is also called as perfectly stirred reactor many times okay and of course that could be abbreviated as PSR and that would that would kind of like rhyme with the next one which is a plug flow reactor abbreviated as PFR so you could think about PSR versus PFR and so on if you do not hear me right then you could get confused between the two so we do not want to get into all that so we will just use WSR okay for the well stirred reactor so or perfectly stirred reactor just use some space below PSR sometimes historically it is also referred to as long will reactor because because of the person who came up with this it turns out apparently that Zelda which actually came up with this 10 years earlier in Russia so whatever whatever you can think of during the time that Zelda which lived you have done it before other people that that is that is Zelda which for you is like a father figure in combustion so here what we are talking about is right so you have an inlet and you have an outlet and so you have a m dot I in in a corresponding Y I in and a H in H I in you can say H I in there is mass specific enthalpy of species I the inlet and correspondingly we now say you have a m dot I out Y I out H I out and so on but all these things of course you now have a control volume that is that you apply within this reactor just pairing the inlet and the outlet okay and consider a Q dot that would that would get out of this reactor with a with the following characteristics so it has a fixed mass flow mass flow rate so that means sigma mi in is equal to sigma mi out equal to I am sorry sigma mi dot in equal to sigma mi dot out equal to sigma m dot okay so you have an m dot we did not use m dot here that is the reason why I was apologetic this is a fixed mass system okay this is a fixed mass flow rate system so when you say fixed mass flow rate implicit in it is that we are looking at steady state you do not have any accumulation of mass or depletion of mass yeah so then course you are now looking at the reactor having species all species with with with concentrations whose spatial variation we would not worry about okay there is a steady state that means we are not looking at any temporary temporal variations as well that means it is not supposed to vary right so if it is not supposed to vary well that means that you do not you do not have to look for differential equations of evolution of these with respect to either space or time okay but they will depend on parameters what parameters typically we are looking at what is the size of the reactor that you want to work with okay and what is the heat out that you want to get out of this reactor so depending upon these and the input conditions you will get the corresponding output or you will now get the composition inside the reactor and the temperature to attain based on what is the size at which you want to operate this reactor size is given by m dot okay that is like the throughput and what is the heat that you want to get out of this so these are now like the key parameters in the problem yeah okay so then of course it has a certain temperature pressure and volume okay and of course you could you could ultimately we will see this there is a there is a relationship between these it is like you are handling a certain m dot at a certain pressure and volume okay that will relate to a certain density all right and with that density for these mass fractions you will now get the partial densities of each of those species to be fixed correspondingly and so on so there are relationships that we will be working out among these okay. So this has a an inlet and outlet outlet of for steady stream steady stream of reactants and products respectively that is reactants get into the inlet and products get out of the outlet now when do you get this kind of situation where you do not worry about this details is when you now have high velocity jets of reactants coming in and then getting into like a tumble or a huge swirl or something of that sort okay so and then finally something that comes out so it is so chaotic inside that you do not want to worry about it right so experimental reactors reactors employing high velocity jets offer this situation ideally okay the interesting thing as we will see is what we are interested in okay we are now looking for algebraic equations without any derivatives right in concentrations and temperature again that those are the ones that we are interested in so the concentrations obviously depend on the reaction rates and the reaction rates depend on what are the reactions so in principle you could actually use a reaction scheme of a large set of reactions detailed elementary reaction steps involving whatever species that you are interested in keeping track off that there are there are going to happen in this reactor on come out of there right and typically this is good for doing relatively quick I say relatively because it is not quick to start with okay you still have like a huge set of algebraic equations to solve which is not easy okay but you are not bothered about the spatial distribution the temporal evolution and all those things you are not really worried about the geometry and all that stuff right so relatively quickly you can get things like pollutant formations like for example you are interested in NOx okay so you can actually get a fairly good handle of what would be the NOx production from pushing in so much amount of air and so much amount of fuel into a reactor that is going to be operating at this particular pressure this bigger reactor and so on okay or maybe carbon monoxide production why carbon monoxide production let us suppose that you are now trying to push lot of fuel in there and maybe also corresponding amount of fog air okay for complete oxidation but this volume is not good big enough for complete oxidation to happen and before complete oxidation can happen you now have products coming out you are going to be primarily looking at incomplete combustion products right that means you are going to have some carbon monoxide that is a pollutant you see so you the volume available is going to also constrain how much reaction can happen and therefore you can you can estimate pollutants based on that as well okay. What we will also see is typically this is used to design reactor sizes like the volumes okay based on another idea of what is called as residence time okay so essentially what happens is you now have a reactor and you want to have a throughput of reactants coming in reacting and producing products and that go out you want to first of all decide what how big this reactor should be right for this you have to have a notion of what is the residence time of these species inside this reactor by the time they could actually go through all the chemical reactions and come out as complete products. So for complete products we will now use the concept of residence time that will be related to the volume okay so that that is another way of doing this the third and very interesting idea that with which you can actually use this this is something like flame blow out alright and how does that work so it is like let us suppose that I have a I have a reactor okay and then I am now trying to pump in more and more and more of the reactants in and at steady state we are now expecting a the same mass flow rate of products out alright and of course for a given volume you are now going to have incomplete combustion progressively but at some stage it is simply not possible for you to combust at all okay as you now throw in more and more reactants at it that makes sense but how are you going to get this out of a model like this the answer is you now have a set of non-linear algebraic equations that you are looking for which may not have a solution at all for certain values of m dot or q dot okay let us suppose that you are trying to extract a lot of heat out of the system as well you can quench right so for the set of parameters that you are employing you may actually get into a no solution region which which now begins to come correspond to a flame blow out situation so yes yes everything is independent time reactions are occurring all the time so whatever you are talking about everything is happening all the time that is a dynamic process okay as you speak right you have now reactants coming in react and produce products or intermediates and products everything is happening so you now we are not in we are not bothered about the spatial variation okay so in reality what might happen is as you come in you have more of reactants here okay more of intermediates here and more of products there as they go out but we are not worried about that that distribution so if you are not worried about that all you have to say is you have in this reactor for a given set of reactant inputs right you are going to get in fact the question that you are asking as to be approximated this way we are looking for this output of products okay but we do not resolve the spatial we do not spatially resolve between somewhere here and somewhere inside okay so in that sense it is like saying it is all perfectly mixed all right and you are now going to get out these species okay so it is we will try to relate to what we did for the adiabatic flame temperature well while we do this okay except we are now going to be able to parametrically vary things here so we do not have a spatial variation we do not have a temporal variation the difference between the adiabatic flame temperature calculations and what we did earlier for the fixed mass systems as well as what we will do later on for the plug flow reactor is we can track variations between the starting point and the ending point which we did not do earlier okay here we are not doing that so the question is for how is it different okay the answer is as far as products are concerned we are going to primarily get something like what we did before and they are like they are not exactly equilibrium products but we will get from similar calculations as we will see okay but we will not do the spatial variation right so in that sense it is difficult for us to understand how it can be steady state okay but in reality this is what we are expecting so okay so no spatial variation no no no I meant to say I meant to write out something that I got distracted from the let us just write out NOx NOx formation incomplete combustion incomplete combustion can be quantified from combustion efficiency okay so you can find out how much of the original fuel that you put in has been completely oxidized okay so and then that denotes what is called as combustion efficiency so that that can quantify incomplete combustion and so you can get that kind of quantity here incomplete combustion and flame blow out okay flame blow out could be steady so considering the complexity of the fluid mechanical process that is happening inside in reality this is a quite powerful idea okay just use only algebraic equations to do this is actually a fairly powerful idea here for these kinds of complicated situations so now let us look at the other thing that we said no spatial variation no spatial variation of concentration or temperature temperature considered and steady state assumption or steady state assumed okay no time variation no time variation as well okay so this should lead to algebraic equations as we will see now so global balance for individual species the reason why I am calling it global mass balance is we are not applying it to any particular point in this reactor it is a it is a it is applied to the entire control volume when we want to derive a mass balance for each species we will actually take a control volume approach and do like something like a global balance but we will keep in mind we will now make use of the fact that the control volume is arbitrarily shaped and then we will be able to notice then that the equation that you derive there is actually valid at any point but here it is not really valid at any point because you do not have a you do not have any spatial distinction at all okay so this is this is still a certainly a global mass balance so at any at steady state at steady state for each species I m dot I out equal to m dot I in plus omega I times V I am sorry make a mistake you will now use wi okay so this is this is like this is a this is actually like a verbal equation okay we just put in symbols here but it is essentially like a intuitive idea you think of your back bank balance for example okay so this is what you had to start with and then you put in some some money you get some more some more back bank balance right so anything anything anything can be budgeted like this effectively right so whatever you get out is whatever you put in plus what you created this now is actually for a particular species if a species is a reactant and it is going to get consumed this is going to be negative so you are going to get less of it when compared to what what went in so strictly speaking we should say this is I equals 1 to n and keep in mind this is a dot here that means these are rates so we are looking at something like kilograms per second okay Wi then is the okay net rate of production of mass of species I per unit volume so long as we were dealing with fixed mass systems we were always handling things on a molar basis okay but the moment we are actually shifting to a open flow system we start dealing with things on a mass basis okay so here instead of using omega I as we did before we use Wi so we shift from Greek to English okay and this is say mass basis and this is what I was trying to point out if you now get a sigma Wi you should look for it to be 0 because mass is neither created or destroyed totally okay but so long as you have a sigma omega I that is all right wherever because number of moles is not necessarily concerned so so look at what is going on this is mass produced per rate of mass produced per unit volume so this should be kgs per second meter cubed so times the volume should get you carry kgs per second okay so these are kgs per second mass flow rates okay. So we know that Wi equal to omega I times capital Wi where capital Wi is the molecular weight molecular weight of species I okay the distinction between my Ws and omegas is you have this little curly thing for Ws whereas that is absent in the omega so you have to have to look for that little thing there okay and m dot I equal to m dot times Wi I and this m dot is the total mass flow rate which is a constant because we have steady state okay and that is a parameter in the problem that tells you what is the throughput we want to handle in this system okay. So if you now want to put all these things here we get omega I capital Wi V plus m dot Wi out minus Wi in Wi out minus Wi in equal to 0 I equals 1 to n okay now you could leave it as it is and then say this is given okay this is a parameter that means you can have a set of Wi I for any m dot okay that means you can send in a particular composition of reactants at any level of mass flow rate you now have a large mass flow rate of the same composition or a small mass flow rate of the same composition Wi in for I equals 1 to n will refer to the composition that is sent in m dot will actually refer to the size how much you are trying to push in through this right. So this is given this is what you are trying to look for okay now since we do not make a distinction between here, here, here and here we do not have a special variation that is possible okay what we are looking for is only mainly what is happening inside as like a whole quantity right and therefore you can say because you have to bring in the volume and so on you now say why I out is the same as whatever is the why I inside the control volume because you just cannot make the distinction okay especially so here we suppose right why why I out is identically the same as why I CV where why I CV stands for the composition in the control volume okay and of course keep in mind this Omega I is going to be a function of concentration of individual species and temperature so temperature is coming into picture here and we do not know that right so for that we need to solve the global energy balance so you get one more equation for that unknown and that is what we should be pursuing okay so global energy balance q dot is equal to i equals 1 to n m dot i out hi t out minus sigma i equals 1 to n m dot i in hi t in okay so let us let us just go back and write why I CV as simply why I let us not have any subscripts anymore for what we want to calculate anything that we want we try to not have subscripts superscripts primes and all that stuff that should be like the main quantity. So here what is happening is this is the product enthalpy enthalpy rate I stand corrected this is the rate at which enthalpy is coming out okay this is the rate at which the enthalpy is going in the reason why we have a rate is because we now have an m dot the dot is what is making it a rate okay and this is beginning to look like what we did earlier okay so when you did the adiabatic flame temperature calculations we were looking at energy not the energy rate right not the power rather so here looking at q dot previously we said q and for adiabatic flame temperature calculations we said we set q is equal to 0 and then equated the product enthalpy not the enthalpy rate equal to the reactant enthalpy okay and then we tried to find out the t out which is what we are going to do now as well alright so it is an algebraic equation that we had earlier which is also same thing as what we are doing here you see except the m dot I out was calculated based on equilibrium calculations earlier on whereas here m dot I out is nothing but m dot times why I out and why I is going to be now calculated with this equation taking into account the chemical kinetics so this is the difference so this is the real process that was a hypothetical process you see so here so you want to now say this is a parameter and this is what we want to try to find out and we suppose t out is equivalent to identically equivalent to TCV which is the temperature in the control volume because we cannot distinguish and that would be equal to C this is given you might argue with me that wait a minute we can't we can't distinguish between here and here as well okay do not do not do not do not ask that question okay that that is essentially the point we are not really doing anything bad strictly speaking this is out that okay this balance is correct we just drop this drop the subscript that is it okay and and so q dot is equal to we can now pull the m dot out sigma I equals 1 to n why I hi of t minus sigma I equals to 1 to n why I in hi of t in so here why I in is also given as we said earlier the base we need to know the inlet composition and recall hi of t is equal to hi of not sorry hi of not I plus integral t ref to t small CP I dp okay so we are writing everything in small it is small h here small h here small CP okay this is because this is on a mass basis the capital CP that we wrote earlier for the fixed mass reactors is on a molar basis okay so we now have two equations and two unknowns we will proceed this a little bit more tomorrow this is this is a good time to stop because we are now looking at the picture okay we will now do some finishing touches tomorrow and then proceed with the plug flow reactor thank you.