 So we are doing two simplifications here one as first of all we are considering one dimension the second thing that you are looking at steady I would like to point out that for what we are trying to do here we do not need to make these assumptions okay one other words the the final result that we are going to get should be valid for a three dimensional unsteady flow as well okay this is just to keep things simple in what we are going to do now what are you going to do now recall in what we did up until now we wrote up the equations of conservation for combustion and found that we had 5n plus 6 equations to solve and n was like 20 40 whatever and so correspondingly we had like a large number of equations to deal with so the question is is it possible for us to a reduce the number of equations and b is it possible for us to decouple some of the equations so that it is sufficient for us to solve only one set of equations for doing something another set of equations for doing something else and these are not exactly coupled with each other except that maybe we get solutions from one and plug into the other and that is it so that means like as far as the second is concerned the first solution is given to it we do not have to solve it coupled with it and as far as the first is concerned we did not have to worry about the second set of equations at all completely and just go ahead and solve for this so this is what is meant by decoupling sets of equations and we will try to understand what this idea leads to in concrete terms or in the words what decouples from what and so on at the end of what we are trying to do but as I said earlier from now on all we are going to do is to completely go in the direction of simplifying things right and then deal with a few simplified situations from now on so one of the way this is not the way of looking at simplified simplification this is the simplification here on one dimension and steady is to just make what we are going to do now simple but at the end of it we are going to come up with some simplification that is valid for multi-dimensional that is three-dimensional steady flows unsteady flows as well so let us now look at the simplified situation if you now have the overall continuity equation overall continuity equation can be integrated can be integrated to give so what the overall continuity equation we had was dou rho by dou t plus divergence of rho v vector is equal to 0 so steady means you do not have a dou rho by dou t 1d means we have a d by dx of rho u okay and if that is equal to 0 we can integrate that to get rho u is equal to constant right so okay the momentum equation the momentum equation for 1d steady flow and let us also neglect body forces now neglecting body forces mean you whenever we are actually making these kinds of assumptions that let us neglect this or assume it to be negligible those kinds of things what it what it we are we are a question it is it okay to neglect body forces okay so in combustion one of the problems that we will find about this is the body force pretty much is the one that is actually giving rise to the shape of the flame as we know it if we now light up a candle okay like for example you go for a candle light dinner party with your date and then you are trying to enjoy the moment unfortunately then you start remembering about combustion because you have this flame and then the entire evening is shattered right so what are you looking there you are now looking for a flame that is kind of elongated vertically okay assuming that your date is happening in a gravitational field and you are not really going in free space so if you did if you do are in free space then or the zero gravity environment you now should actually look for like a bulb shaped flame whereas if you now are in a gravitational field you have a buoyancy force that is actually trying to push this flame in an elongated fashion so the question is it okay obviously if you neglected body forces you are not going to get flame shapes that look like anything that we are used to around here on this planet right well the answer is maybe all that shape is not really important okay for the purpose of burning the candle it is a heat that is actually being released closest to the candle surface that is really what matters for melting the wax and getting the liquid to go through the capillary pores of the wick and then vaporizing to feed to the flame so we did not have to really worry about the shape of the outer shape of the flame far away from the wax surface the candle surface right so many times what you will find is it is okay to neglect the body force not a not a big problem there so if you now did that then you get your momentum equation which looked pretty ominous previously it is now going to look like u du by dx equals 1 over rho d by dx of p plus 2 over 3 mu du over dx minus 2 mu du over dx well as we can see from what we have done even the continued equation the there is only one direction that we are looking at the x direction and we are only looking at one component of velocity that assuming it to survive which is the x component of velocity and that is called you okay and since you are having looking you are looking at only one direction it is sufficient for us to actually replace partial derivatives by ordinary derivatives with respect to x so you have a u du by dx is coming from the convective term is equal to minus 1 over rho d by dx of p that is a pressure gradient term and then this is the bulk viscosity term and this is the viscous shear stress term okay so these are the terms that survived the unsteady term dropped out because we are looking at a steady flow pretty much all of the terms are around of course the body force has been neglected okay so this could now be written as if you now write this well what would I like to do if it is possible for me to write the whole thing as a derivative what did we get over here we had a divergence of rho b is equal to 0 for the vector form in 1d we had a d by dx of rho u is equal to 0 so similarly if I can actually get something like a d by dx of something equal to 0 I should now be able to say that something as a constant that is very easy for me to integrate it becomes like a exact integral for me to evaluate therefore so we can now write this as let us say d by d by dx of rho u well rho u square okay equal to minus d by dx of p plus 2 by 3 mu du over dx minus 2 mu du over dx how did I get this well first of all I could take this row to this side okay and notice that rho is rho u is equal to constant so if since rho u is equal to constant I could actually take it inside the derivative all right and there is a reason why you why I got a d by dx of rho u square okay. So what that means is I had a d by dx of this equal to minus d by dx of that which means I can combine both and have 1 d by dx right so if I now put everything together and have a d by dx equal to 0 then whatever is inside the d by dx should be equal to constant which is now in our case rho u square plus p minus 4 by 3 mu du by dx equal to a constant okay. We are now beginning to see things that we are somewhat familiar with from let us say gas dynamics previously in gas dynamics you would have done typically inviscid flows where you had a p plus rho u square is equal to a constant for compressible flows okay in fact rho u is equal to a constant actually comes for compressible flows right otherwise you would have had something like a of course in quasi one-dimensional situation you would have a rho ua is equal to constant this is just one-dimensional not even quasi one-dimensional so you do not you do not worry about an area you only worry about a mass flux you do not worry about a mass flow rate okay but you have to take in account the variation of rho with velocity that means rho is varying it is a compressible flow okay and there is a reason why you have a rho u square plus p okay but that is not equal to constant like what you had seen for inviscid flows because we are still considering viscosity so you have this extra term that comes into picture you are a viscous effect okay and that is actually putting the bulk viscosity and the viscous shear stress terms together you get a minus 4 4 by 3 mu du by dx so previously we did not have this term we now have we may now have this if you were to look at like a incompressible flow you need to have a half rho u square plus p is equal to a constant for Bernoulli's equation the half did not come because rho was varying here okay previously if you had only a incompressible flow the rho would not vary then you would have a ua is equal to a constant for a quasi one-d that is what you are familiar with for a Bernoulli's equation approach there so all these things are not very different from what you are what we have done with done before maybe with the exception of having this extra term yeah long later when we are trying to do detonation we will try to revisit the situation and then see how the flow transitions from its initial conditions to what is called as a Chapman-Jugey detonation later on through many states that correspond to this okay so what we will want to do now is let us say we can write this as rho u squared over p plus 1 minus 4 over 3 mu over p du over dx divided by sorry equal to constant over p why am I doing this the answer is I want to have a one in this so that I can compare the other terms relative to that okay it is sort of like always looking at making a comparison with 100% so how many marks is this going to score with respect to 100% how many marks is this going to score with respect to 100% okay 100 so 1 essentially is like a benchmark for us so why did we actually choose pressure to be the one of the denominator and make the pressure term equal to 1 the answer is we are now going to actually examine how this is going to compare with 1 and see that under low Mach number conditions this is very small and this is also very small all right so all small with respect to what small comparison to what with respect to 1 all right so that is what we are trying to look for so p note that p over rho equal to RT a squared equals gamma RT where a square a is the speed of sound so this this implies that p over rho equals a squared over gamma or you could have said a squared a squared is equal to gamma p by rho right which is something that we know all this is for a perfect gas of course therefore then if you now try to plug p by rho p by rho would now be in the denominator in this term right so u squared divided by p by rho is what we have for this term and that would be simply u squared divided by a squared divided by gamma which is something which is basically gamma m squared right so u squared over p by rho is equal to gamma m squared okay I keep this with you let us look at the third term over there the second term of course is 1 so let us look at the third term from kinetic theory I told you that we have to have something something for mu and mu is a is a transport property and transport properties can be evaluated only if you go down to molecular level that means you have to go down to kinetic theory or we cannot get it from continuum effect and within the continuum framework you simply have like a law that states that shear stress is directly proportional to velocity gradient and the constant of proportionality is something called viscosity okay that is like a law that we just observe in the continuum world without knowing why if you want to know why you have to get down to the molecular level and then you will be able to find out what viscosity actually means in terms of molecular level fluid motion all right so if you want to know how the mu depends on things like pressure and density and temperature and all those things typically you can try to relate it to two thermodynamic two thermodynamic properties and then there is a dependence okay so what you are looking for is mainly like a dependence rather than the exact expression because you are looking at an order of magnitude approach okay relative to unity we are trying to see how this pairs relative to unity so mu can be written as rho a lambda where of course a continues to be the speed of sound now how did you get the speed of sound to come in for the viscosity that is not very surprising because if you want to actually try to find out what is speed of sound you have to get down to the molecular level as well because it is a molecular level phenomena that is actually propagating sound all right so all these things are related there lambda is a mean free path what is mean free path mean free path is basically the average distance between average distance covered by a molecule between successive collisions all right so now what we have to look for is du over dx that means we have to look at an order of magnitude of what that derivative is going to look like so if if delta u is of the order of u that means the changes in velocity or of the same order as the velocity itself as in this is not to say that the changes are large it is essentially saying that the changes are actually measured as appreciable fractions of this of the velocity so for example if the velocity is let us say 10 per second and when you are now looking at du by dx in fact we should now so also start looking at however what is x going to vary like so delta x let us suppose it goes as L L is like a characteristic length characteristic dimension of the problem so you we come across these kinds of things when you do things like flow past a sphere or something or flow through a pipe and so on so where we have characteristic dimensions right like they like this fear diameter or the cylinder diameter or the pipe diameter all these things these are like characteristic dimensions so if you now have flow through a pipe and the pipe is about let us say 50 millimetres large then you would try to actually measure the or take divisions like go 1 mm apart or something of that sort right so what we are talking about as x is it is going to be a fraction of the characteristic dimension or in other words we are not going to take delta x to be very much smaller when compared to the characteristic dimension okay it is going to be of the same order so if you now have like 50 mm we are going to go in steps of like 1 mm which is of the order of about point 0.2 okay. So that there is still all like in terms of what the characteristic dimensions are if for this kind of change in X okay of course X is in the axial direction in this in this particular case so if you want to now look at what happens in this direction you could even do much larger than 0.1 mm you could even do things like this let us look at what happens every 1 every 5 cm 5 mm okay so delta x down simply becomes like 0.1 which is which is 0.2 so you can take larger steps and if velocity is changing if it is going at 10 meters per second and we are changing by 1 meter per second over a let us say 10 mm difference then these are all d du over dx is of the order of u by L this is essentially what you are talking about what you are saying okay. So then 4 by 3 mu over p du by dx now goes as let us forget about 4 by 3 4 by 3 is not very different from 1 okay now this is this is some new math so where we kind of cheat on these like course when you are in your fourth standard or something like that you really start mugging about these decimal places and all that stuff and then as you grow older and older you become like lot more scrupulous about like 8 decimal places accuracy and so on and then finally you come to graduate school and we say like 1.33 is the same as 1 forget about these two decimal places and all come on is this why we studied so hard okay that is exactly what is meant by order of magnitude now so let us not worry about the 4 by 3. So we will now say or if you want to worry about it keep it don't worry about keeping it or leaving so then mu goes as 4a sorry 0 a lambda then keep it as p here and as I just try to believe at the point delta u over delta x is like u over L right and we should now know how to rearrange this you can get the get the row down here so as p by row and then we now notice that p by row is a square by a square by gamma so I cannot plug in a square by gamma over here without the row there and then a gets cancelled okay the gamma goes to the top and so then you have a a over here at the bottom there is no a at the top anymore but you and a get together to make an M and the lambda and L get together to make what where is the last time we thought about the mean free path the Knitzel number right so the mean free path relative to the characteristic dimensions of your problem okay is something that you come across the first time when you do fluid mechanics typically when you are you are you are convinced or rather brainwashed into thinking that we will have a continuum approach that can be justified by looking at something called a Knitzel number which is very very small for typical situations like the kind of flow that we have in this room okay atmospheric pressure room temperature the air is so densely packed when compared to the mean free path let us say you are not looking at flow past this table okay that is like billions of mean free paths there okay so the Knitzel number so this this together this gives you the Knitzel number and as I said you can get a gamma gamma M from the other part so so this is actually going to give you a gamma M Knitzel right if you want me to write what this is that is Knitzel number KN right and for typical continuum applications KN is very small okay KN is like 0.001 0.0001 whatever so very very small number 10 to the 10 to the minus 3 okay whatever so now KN is very small for continuum flows right then the question that you ask is very minute we are doing combustion is combustion continuum flows is why we did not start talking about like molecules colliding with each other and reacting and then you still even talked about like the molecules need to be exactly oriented so that you have like effective collisions and then you will have the chemical reactions for some part some atoms in some parts of the molecule with some other atoms in some other parts of the molecule and all that stuff when we talking about all those things at the molecular level why we now suddenly saying continuum yeah I mean continuum does not mean that there are molecules okay so the fact that we have to actually use the law of mass action and the Arrhenius law is because we are actually looking for a continuum representation of those chemical reactions the fact that we are able to actually define a concentration of a species at a particular point which actually contains billions of molecules of all the species at that point okay means that we are actually dealing with continuum so even though the chemical reactions are happening at the molecular level it is happening at a point where you are looking at millions of billions of molecules of lots of species at the point with each of them being able to be represented by a concentration at that point concentration means you know to have you need to have like certain mass for that given volume of that little point or a certain number of moles of that for a given volume at the particular point okay so you are now saying that that point has a little volume and then there is a mass associated with it and all those things and the point that we are talking about is like a mathematical point in space where your chemical reactions are happening now it is not like it is happening in this room or things are happening the size of this room you have to start now imagining like a flame that you are in that you encounter okay let us say for example if you have a Boonson burner you now have a flame you know thinking about a flame that is like a millimeter thick or even smaller okay and then you now should be able to pick a point within that flame and at that particular point you now have all these chemical species with their concentrations and so on all right so it is it is a continuum flow that is not a problem at all and therefore so finally what do we have for low mark number for low mark number flow how low how low is low 0.3 0.3 is low no 0.3 is like this magical number that we have in our minds about when the flow is still come incompressible or just become and I don't worry about all those things right what we are talking about is mark numbers that are even lower right that they are like about 0.1 or less very 0.01 or 0.001 okay we are typically talking about hot flows here the temperature is large there the speed of sound is very large okay therefore I should say the other way that the temperature is very large therefore the speed of sound is large because it goes a squirt of temperature so you have a lot speed of sound so even for a moderate velocity okay your mark numbers are fairly fairly small so if you were to think about a mock number of about 0.1 I think I made this point before if you now think about a mock number of 0.1 that is happening in like a let us say a gas turbine combustor okay and the temperatures there are in the reaction zone of the order of like 2500 Kelvin therefore the speed of sound is of the order of 1000 meters per second then we are still talking about velocities that are about 100 meters per second that is like this pretty the flow is fast for the mock number is low okay so we are not talking about low mock numbers less than 0.1 okay and that is not a magic number it is not like a God given number if it is like 0.11 then do not start shivering huh which is like 0.09 do not feel like like a huge relief for something of that sort nothing of that sort it is very low that is what you are saying is it is quite low when compared to 1 and then what do we have we now say this term which is now beginning to look like this if m were like 0.1 m squared is like 0.01 okay and Lutzen is already like 0.01 or less and m is again 0.1 so this term is going to be much smaller right so then both rho u squared over p, 403 u over p du over dx are much smaller when compared to 1 that means in this equation we should now try to get rid of this and this and then we are left with only one okay we got rid of these because they were actually very small when compared to 1 so if you now get rid of these we have to keep the one okay so we now keep one and then what do you have the right hand side we now say 1 is equal to 1 is of the order of constant divided by pressure that is to simply say that pressure is approximately a constant okay so this this simply means that p is approximately a constant so what are we done we have practically shattered the momentum equation down to a p equal to constant equation we started out with a momentum equation and then just systematically dismantle the momentum empire if you will okay to get it down to a teeny-weeny equation that is like just p is approximately equal to a constant is it okay does it make sense so you light up a Boonson flame do you feel like a pressure around okay so you go to your kitchen gas stove and then light up the gas stove do you feel like a pressure around you do not really get like a lot of pressure around okay it is just approximately constant atmospheric pressure around there that is all right but what about a gas turbine that is really going at very high speeds I am pretty sure there must be a lot of pressure yeah the pressure is high but it is approximately a constant right so we are talking about fairly low subsonic combustion happening roughly at a constant pressure all right as a matter of fact whenever you are looking at these kinds of things like you know neglecting body forces or when you are now saying let us say continuum and then we are sorry now looking for low Mach numbers we do not necessarily have to think combustion right if you now had a non-reacting single species flow non-reacting single species flow okay like ordinary flow mechanics and we now say let us not worry about body forces that is okay in fact in combustion that we had to worry about body forces and then neglected right then it is its continuum flow sure so Knitzen number is quite small and we are thinking about low Mach number flow mechanics there anyway so all these all these assumptions that we are making along the way are satisfied for a normal flow mechanics flow and then we should now be able to reduce our momentum equation down to P is equal to a constant why are we stuck with a momentum equation that we are not able to solve for 300 years when it should just be said as P is equal to constant are we are we getting in any information here the answer is to think about it you just have a piece of pipe let us say a small piece of pipe and then you now blow through this pipe okay and then of course you can play with this and again you can put your hand at the other side and then feel the cool air that is coming out because you are blowing and so on of course you do not really get a lot of pressure out there so it is all approximately atmospheric pressure that is really the key how much did you blow with right the answer is a little bit more than the atmospheric pressure therefore you could actually get the flow out to atmospheric pressure so the delta P that we are talking about there is very very small so that this flow field is approximately at a constant pressure where there was a small pressure gradient that was required to drive the flow this is not this is what we are talking about for this kind of low Mach number flows okay so this is not to say that the pressure gradient does not exist it is quite small and that is good enough to actually drive a low Mach number flow as a matter of fact in a gas turbine with all the complicated machinery like you have these furloughs and you have a flower and a sudden expansion and all these things you are expected not to have a pressure drop of not more than 5% if you have a gas turbine that is having a pressure drop of more than 5% it is a bad design okay you try to try to get it up to 95% pressure recovery all right so you are not looking at a very huge change in the pressure at all therefore for all this big flow that we are talking about okay so this for us does not mean that the momentum equation is gone the momentum equation is still required for you to get the flow field except if the momentum equation now gets decoupled from the species equation and the energy equation this is what it really means okay that means you are starting out of the overall continuity and the momentum together they constitute what is called as the flow problem if you now had the flow solve for that is basically cold flow so you are given a very complicated geometry with all this let us say a pre diffuser a dump diffuser which is very highly curved and then you have the swirler and atomizer that is coming through and then you have you have these liner pores and then you have flare and then all the secondary holes and then the dilution holes and then you have this liner flow and then there is a casing around and you have this combustion going on inside all these complicated things don't worry about all that stuff just take this geometry and then do a cold flow okay work out the flow field in this that means you now take your incompressible equation set for flow mechanics namely the continued equation and the momentum equation solve for the flow field okay you get your velocities these velocities now can be plugged in in your species equation and your energy equation that means we now want to worry about the combustion problem as constituting as constituted by only the species equation and the energy equation. In other words we now say we want to split the set of equations that we have had into a flow problem that you solve independent of the combustion problem and you solve this flow problem get the flow field plug it into the combustion problem so what is the flow problem overall continuity and momentum okay what is the combustion problem species conservation and energy these are now decoupled and there is only one way interaction which is you solve the flow problem independent of the combustion problem get your velocity and plug it in the species where do you read this in the convective terms for the species mass fraction convection and the enthalpy convection in the energy equation so you need the flow the mixture velocity right so you now plug this in there right and then this is your combustion problem you just solve this set of equations for a given flow field okay now many times the combustion people do not have time for the flow mechanics people they get just assume a flow field okay I mean if you really strict enough you would probably assume a flow field that satisfies overall continuity and momentum and many times in simplified problems a assumed flow field that satisfies overall continuity and momentum is simply a uniform flow okay so you just assume a uniform flow that naturally satisfies overall continuity and momentum and then plug that in into your species conservation and energy conservation okay and it also simplifies your species and energy conservation because in our looking at a V.del yi and V.del t kind of terms right and then so if you now look at V.del if you have only one dimensional velocity that means velocity in only one direction it simply reduces to you do you do why by do yi by do x and you do t by do x you do not have a V term at all okay so the lateral convection term gets absent so you do not have to worry about it okay. So this simply means that we do not solve the overall continuity and momentum equations as part of the combustion problem right obtain the flow field by solving them and use it as prescribed in the combustion problem right so what is meant by the flow problem then is overall continuity plus momentum equation right and then we have the combustion problem which is the species conservation and energy equation. So when you now say this arrow that means flow field prescribed so many times that means we do not even care to solve the momentum equation or the flow problem you just take a flow field that is prescribed as far as the combustion problem is concerned now I do not know how many of you are really thinking about this as well as we should what do you mean by saying I prescribed the flow field and then the convection of species and enthalpy is going to happen as prescribed right the question that you have to ask is does this affect the flow field back should we have an arrow backwards well why I would like to say two things here one if you want to have an arrow backwards still be happy that I could draw these boxes separately and identify a flow problem different from the combustion problem and then have a interaction that is explicitly understood between these two as opposed to putting one big box for all the equations and saying that is it you have to solve all of them all the time okay I am still thinking about decoupling these problems that means my solution procedure for this may be different from the solution procedure for this this is the one that is containing all these bad terms like the chemical reaction terms and all those things so I may have to have a solution procedure that is different for this than this right so it still makes sense for me to think about these two distinct problems even if they have a two-way interaction but what I have now talked about is only one way interaction okay but shouldn't you really have a two-way interaction that means shouldn't the combustion actually affect the flow field why would it yeah of course it will we now think about a candle flame right so now let us light up a candle and then assuming that we are thick skinned which I am okay so I now take my finger very close to the flame but somewhere near the bottom of let us suppose with this with this the flame okay it is a big it is a it is a what you call exaggerated view here right is the flame isn't so big but I now take my flame somewhere here okay of course I am beginning to feel the heat all right but I do not really sense a draft I do not see any like I do not sense a flow okay but when I now try to take my hand somewhere there do I sense something now see a flow you have a flow of course you might claim yeah this because of the buoyancy oh hot gases are rising up all right I can claim what if I were to actually go into free space or in zero gravity I should now have a flow that is set up readily outward okay instead of going up because we are not in the gravitational field all right but still I am going to have a flow which is not the same as having more like still still are at the bottom okay that is to say if you now have a flow that is going through a flame for the reactants okay the density is low sorry that the temperature is low so the density is high okay but then when it now gets past the flame and it becomes products the temperature is now high so the density of this for more or less constant pressure right is now going to be low so what does it mean for a low density as far as the velocity is concerned if the row u is equal to constant the velocity is now going to go up so sure enough if you now were to think about like a Bunsen flame right now have a cold flow that is coming in this is like a blue flame all right but then the flow actually enlarges and goes up like that faster than it came in into this flame so the density effect on temperature sorry the temperature opposite the temperature effect on density has the effect of dilating the flow once the flow goes through the flame which in turn means that obviously the flame affects the flow or the combustion problem affects the flow problem through the density depending upon temperature okay so you do have a in effect okay this is because row goes as 1 over T I say 1 over T because we still assume or suppose that P is approximately a constant right P equals row RT P is more or less constant therefore row should actually decrease just as well as the temperature increases so to give you give you an idea of what these numbers are if you now think about a reactant that is entering a flame zone or combustion zone at like say 300 K all right and then it gets out there at let us say something something that is divisible by 3 is to make our life simpler 2400 K okay that is that is more like it okay course security rate you could have taken 2100 K or 3000 all right but we are now talking about a difference a ratio of about 8 anywhere between 7 to 8 to 9 something right so you know looking at a ratio for about 8 times jump in temperature for typical hydrocarbons and therefore the density also should actually decrease the stove so much so much more and then the velocity then increases so much more right so but the Mach number does not change a whole lot so the velocity increases but the more because the temperature came down so the temperature got up therefore the Mach number came down okay so that is not a problem you are still going to have like a P approximately equal to constant but you now have a high velocity there so it is possible for us to actually take this into account and then go back and then alter the flow and then prescribe a new velocity field into the combustion problem find out how the combustion changes the temperature field and then therefore the density field and therefore the flow field and so on but still this means we can actually decouple the two problems and solve them with different strategies okay but take an account this coupling