 So, we stopped here last class where we derived the energy equation across the discontinuity representing the reaction zone and we wanted to see how this looks like in what is now beginning to shape up as goth's called as a Hugonio plot. So this is what you would actually call okay to some extent a simplified Rankine Hugonio equation. So how do you look at this curve in the Hugonio plot, the Hugonio plot is a domain of P infinity but 1 over rho infinity and then you begin to look at how this looks like you now have a coefficient ? over ? – 1 times P infinity divided by rho infinity. So this is itself like a product of P infinity and 1 over rho infinity right and of course we now say P not over rho not is like a parameter that is represents like the initial state and again you can look at a P infinity times 1 over rho infinity coming up over here with a different coefficient of course therefore it is not going to cancel typically you will always look for things like things getting cancelled but that is not going to happen here and of course you now have a mixture of things like P infinity over rho not and P not over times 1 over rho infinity times again you have a P not over rho not term. So this is beginning to look like if you are now looking at like an XY plane you are now beginning to look at a curve that is kind of like some AXY plus BX plus CY plus D equal to 0 right. So if you now have like a leading order term like XY what would that actually signify if XY equal to 1 if that is what you had okay that is like A is equal to 1 and D is equal to – 1 then you would simply and B and C are 0 in whatever format that I just template that I just explain you would be looking at a rectangular hyperbola but since you also have a P infinity divided by rho not term and a P not divided by rho infinity term we also have some B's and C's that are non-zero therefore it is not exactly rectangular hyperbola but it is going to be modified so it is going to be like a hyperbola here. So hyperbola is going to now look like if you now had something like that. Now we do not know exactly if it is going to be a rectangular hyperbola in this P infinity 1 over rho infinity X domain it is very likely that you could have this cross over to lower value okay. So I am not saying that this exactly is the curve at the moment but what we are trying to do is to sort of construct this curve as we go along okay well why did we want this curve because we wanted to solve these conservation equations across this continuity to locate react product properties given the reactant properties including U0 which is something that we are not very sure that we know to begin with and U0 is actually embedded in the slope of the Rayleigh line but the Hugonio so this is what is now called the so this is rectangular hyperbola is what is called as a Hugonio curve and the Hugonio does not involve any m dot in other words it is like a purely thermodynamic curve okay. So the flow information is actually buried in the Rayleigh line which is a mixture which is a combination of continuity and momentum together so that carries the flow information whereas this the energy the thermal energy equation effectively is not covering any flow information there and then we are what we are looking for is a intersection of these two the these two curves that is a Rayleigh line and the Hugonio curve and sure enough we find that there are a couple of intersection points which will now tell us that if you now started out with the P0 and 1 over rho 0 as the reactant conditions and this particular point is typically referred to as what is called as the origin of the Hugonio plot and then what we want to see is where is actually the ma the conservation equations where are they satisfied again right so what you will find is that you now get like about two points where they could be satisfied so you have like at least two possible solutions this is a great progress that we have done starting from when we did not do when we did not have any equations to solve we had like an entire plane to search the solution and then when we now had the Rayleigh line we now reduce the possible set of points to a line rather than a plane and since you now also have the energy equation in the form of the Hugonio curve we look for point point of intersection of course you are now beginning to see like about two points of intersection instead of just one right so we now have this look at what are the different possibilities for these two points of intersection whether it be only two or whether it be whenever it be one and if it is two what are the possible two values and so on right obviously this is a this is a curve where which has the same same sign for the curvature that means it is always like a like concave facing upwards and side upwards to the right therefore since the curvature does not change sign you do not have any points of inflection in the Hugonio curve when a straight line intersects with this you can have up to two intersection points you are not going to expect more than that right but you could look for one intersection point or none when would you get one intersection point if the Ray if the Rayleigh line were tangent to the Hugonio right or the Rayleigh line never really intersects with the Hugonio then you do not have any solutions right so there is some hope we are now looking at zero solutions one solution or two solutions not more than that good so with this picture we now see how to go about constructing this plot little bit more carefully right the first thing that we notice is the first thing that we notice is if your Q is greater than 0 okay so for Q greater than 0 the Hugonio curve passes to the right and above the origin that is there okay for Q equal to 0 you could also check this if you want by plugging in values for P infinity and rho infinity for Q equal to 0 the Hugonio curve would pass through the origin you see right now what is Q Q is essentially negative of what is called as H infinity not minus H not not and H infinity not and H not not or in turn sigma i equals 1 to n delta H of i yi not delta H of i not and sigma i equals 1 to n yi infinity H a delta H of i not right so it is only because of the compositions at yi at the not condition and the infinity condition the delta H of i not is the same for both because it is a standard heat of formation right so we are now getting these different hi not and hi infinity and then the difference between the two is essentially the chemical heat release with a negative sign okay so that is a chemical heat release if you Q is equal to 0 that means you are not having any chemical heat release so the chemical heat release is not there essentially we are talking about a non-reacting mixture right if you have a non-reacting mixture with a Hugonio curve that is somewhat similar to what you might have gone through in a basic gas dynamics course with the where you are still looking at as a matter of fact if you now think about how we had this equation we had like a H not plus half u not square is equal to H infinity half u infinity square that is exact without the Q right and that is exactly the same as what you would call as an adiabatic energy equation that you would that you would write for a energy conservation across a shock okay so across a shock you would find that the stagnation temperature remains the same the stagnation enthalpy particularly remains the same right that that is true when you do not have any heat release okay or any chemical heat release. So a curve that actually passes through the origin with Q equal to 0 is what is called as the shock Hugonio right now that tells us that if you now have a Q not equal to 0 and we expect typically the Q to be greater than 0 rather than less than 0 we are not looking at cooling kind of thing right so what is it when you now have a wave that is reacting in addition to being a shock right so you are now beginning to think away and then of course a shock is actually either facing a supersonic flow or it is traveling at supersonic flows this is supersonic speeds relative to like still reactance what is it I mean this is not something that we thought about we started talking about something like low Mach number conditions and all those things but now we are getting into something but of course in what we are doing now we are not really making the low Mach number assumption right we should be now game for any kind of Mach numbers that are approaching your wave but the language that we are beginning to use is now beginning to now admit the possibility of supersonic flows right okay when would that happen so let us just do this a little bit more carefully let me let me declutter this picture of all the terminology that we have that we are now used to and we just want to now redraw this only with things that we want to now take take so let us say you now have the origin again I am not going to name it now and let me first draw the Hugonio line a little bit farther away and let us now say that you go here and here and let us see what happens here your Rayleigh line can have only a negative slope right so if you now if this is your origin of the Hugonio the Rayleigh line can fill only the third quadrant sorry the second quadrant or the fourth quadrant right it cannot go like this so you cannot get into the first quadrant you cannot get into the third quadrant so this part of the Hugonio is never going to be used right so we now have to erase this part away from consideration we just do not want to worry about that part of the Hugonio at all we will worry about a part of the Hugonio that starts from where whatever rho infinity is equal to 1 over rho 0 okay another part of the Hugonio where P infinity is equal to P 0 what does that mean if you were to now have a solution that is somewhere here that means your Rayleigh line is sort of passing like this it means two things one your final pressure is about the same as the initial pressure okay the pressure hardly changed when the flow went past this wave second this Rayleigh line has a almost like a zero slope that means the wave is not moving much so it is possible that if you are talking about nothing yes it is like you do not have a wave so obviously the pressure will increase yeah that cannot be true because the density increase the density decreased so there was a density decrease that was primarily because if the pressure remains the same and the density decreased that is because the temperature should increase so you had a heat addition all right that is there is a reason why the Hugonio is shifted away from the origin in the first place so you had a heat addition correspondingly you had a temperature rise but you do not really have a wave that was moving fast enough and you did not have a corresponding pressure rise this is now beginning to look like what we were talking about for what is called as low Mach number conditions right the pressure is approximately a constant right so low Mach number simply means that the the wave is not moving too fast or relative to the wave the reactants are not moving in too fast right therefore the Mach number is very low so you now have one branch of the Hugonio that is now beginning to correspond to waves that are moving kind of very slowly okay and the fastest wave could possibly have a slope that is tangent to this at this point that is the fastest wave that is possible that is the highest m dot that this branch of your Hugonio will intersect okay if you for any faster waves the Hugo the Rayleigh line is now going to actually go downward and fail to intersect the this branch of the Hugonio will that intersect the other branch not not not immediately you now see that you have to actually go to a still steeper Rayleigh line that will begin to intersect the other side or if you are thinking about an experiment where you want to conduct a stabilization of a wave by sending in reactants at faster and faster flow rates right between this particular flow rate and that particular flow rate you just cannot stabilize a flame at all and then you again begin to stabilize flames for all Rayleigh lines that are now having steeper slopes than this you can have points of intersection until you get to this point which has an infinite slope right it is kind of again just like how a 0 slope is almost impossible and we are thinking about like very low velocities right infinite slope again is not something that we that we that we want to think about but very high very essentially very high slopes that means very high velocities for which the pressure jumps but the volume did not change a lot so this is more like a isochoric process that is a what you started out with was the isobaric process right so in the one extreme you started out with an isobaric process corresponding to very slow waves right and then you could think about a this little bit faster waves but with a certain decrease in pressure not a whole lot of decrease okay from here to here you see that the pressure is decreasing only a little bit okay but the expansion that is the density decreases as you go this way 1 over density increases so density actually decreases there is a significant amount of expansion that is going on right from the beginning for because of the heat addition and then nothing until you get to very fast waves and you can now get these very fast waves to go all the way to the other limit where you are getting close to isochoric process that is like a constant volume situation right and there again because of the heat addition you hardly have any change in volume but essentially the pressure increased right so what you then have is this is what is called as the lower branch of the Higoneo and this is what is called as the upper branch and we want to call these points the detangents the tangent relays intersecting at these points as the LCJ point and this is the UCJ point Cj stands for Chapman Juge so the LCJ indicates the highest velocity of what is called as deflagration waves so this is the lower branch corresponds to what is now called deflagration right. So LCJ the fastest deflagration wave and the upper branch of the Higoneo corresponds to what is called as detonation and UCJ corresponds to the slowest detonation wave right and the m dot UCJ is still significantly greater than m dot LCJ that means the slowest detonation wave is still much faster than the fastest deflagration wave that simply means that you now have two classes of two classes of waves that are significantly apart there is really no overlap at all in their wave speeds and in fact what I would actually what I should actually say is this would be much greater than so in terms of wave speeds like instead of writing m dots if I were to write you not assuming like they start out with the same row not right what this now and see I told you that I do not even I do not know what the u0 is right I am still not going to know the u0 until the end of this exercise but I am beginning to get some ideas about what the bounds of my u0 should be for two different kinds of processes that we are now beginning to think about namely deflagration and detonation right so what we are talking about here is for detonation for deflagration waves your u0s of the order of a few tens of centimeters per second under laminar conditions but your detonation waves so to a few tens of centimeters per second is still less than a meter per second but your detonation waves are typically of the order of a few kilometers per second so now see that there is like about three orders my three to four orders magnitude change difference a factor of three three to four orders magnitude difference between the detonation velocities and deflagration velocities right what we should further notice then is that we want to think about the deflagration is going as subsonic waves where as the detonations are going at supersonic waves but before we do that we want to now quickly go through some mathematical properties because we started out with mathematical equations for these like this one and then we notice that that that goes like a hyperbola and then we of course had a simpler mathematical expression representing a straight line very very easy to see this for the Rayleigh line and we are now looking at these intersections and so on what I would like to first of all before we proceed further is to give you expressions in these are not very difficult for you to derive and these are all like typical exam exercises okay so you can figure out for example depending upon the Q what is this point okay the lowest value of 1 over rho infinity possible for the detonation branch and where is this point for example what are the coordinates of your LCJ point that will also depend on Q and what are the maximum so what is the maximum value of 1 over rho infinity that is possible right and what are the maximum values of P0 that is possible that is of course equal to P infinity and what is the minimum value of P infinity that is possible what you will find is you will you will notice if you now try to actually look at the mathematics of this this would ultimately go and intersect the x-axis for the abscissa right and where it intersects is where you would get the maximum value of 1 over rho infinity and the corresponding value of P infinity would be 0 okay as a matter of fact what you will find is this rectangular hyperbola asymptotes to a value of P infinity that is negative so what you can actually do is to go here and find out the horizontal asymptote for this rectangular hyperbola and you should find that that is negative and since you cannot admit negative pressures in reality you do not worry about where exactly it asymptotes and you have to stop thinking about it at where it hits the abscissa that is corresponding to your P infinity equal to 0 and then there is like a corresponding 1 over rho infinity that is a maximum value to which the proper the products can expand right. Similarly you can look at the upper branch and find out that the lowest value of P infinity it can take will depend on Q but the corresponding 1 over rho infinity will be equal to 1 over rho 0 right then it can also locate this point the ucj ucj and then where does it go so what you will have to find out is there is a vertical asymptote which is having a positive value of 1 over rho infinity that is the lowest 1 over rho infinity that it can take but it is now going to asymptote to that that means the maximum value of P infinity that can take is infinity when it hits a asymptote eventually at infinity right. So these now give you limits for these pressures and densities that it can take in these different branches of the Hugoniot curve so we can go through that but I think even before that we can point out that so in the in the deflagration branch in the deflagration branch P infinity is less than or less than or equal to P not right and 1 over rho infinity is greater than it is certainly greater it is not greater than equal to because it would be greater than or equal it would be equal to 1 over rho not only for q equal to 0 for positive values of q you can never get your 1 over rho infinity to be equal to 1 over rho not so it is always greater than 1 over rho not for q greater than 0 that is that is rho infinity is less than rho not the density decreases and the pressure decreases okay for q greater than 0 you always have T infinity greater than T not this is heat addition so the temperature increases right so with the temperature increasing density decreasing is not news but pressure decreasing these two together then given this means you now have the products go through an expansion right so it is when you have an expansion is when your pressure decreases and your density decreases even with the temperature increase right so a deflagration branch corresponds to an expansion wave right on the other hand if you now look at the detonation wave in the detonation branch your P infinity P infinity can never be equal to P not for q greater than 0 right only for only only for q equal to 0 will the he going to pass through the origin and you have only one tangent relay right but for q greater than 0 P infinity is always greater than P not for q greater than 0 which implies again T infinity is greater than P not you always have a heat addition that you are having in mind in both cases right T infinity is greater than P T not all right so P infinity is always greater than P not and 1 over rho infinity can be less than or equal to 1 over rho not right so the highest value of 1 over rho infinity that you can hope for is this right that is equal to 1 over rho not all of the 1 over rho infinities are going to be less than that and this implies that is rho infinity can be greater than or equal to rho not that means you now have a density increase corresponding to a pressure increase accompanying a temperature increase when the temperature increases you would like to think that the density should actually decrease but if the density decreased sorry if the density increased accompanying a temperature increase that simply means that the pressure increased a lot more right sure sure enough you look at the way the curve goes it is going crazily in the along the pressure axis right so the pressure is obviously increasing a lot so taken taken together all these things mean that we now have a compression wave right so a detonation basically corresponds to a compression wave so we talked about a shock you go Neo we are now beginning to look at one of the branches actually corresponding to compression so it is all kind of going together you can see that there are elements of what we have done before in gas dynamics beginning to look like a special case of what we are what we are talking about right it is a special case in two ways one first of all it is a non-reactive case okay what we went through in gas dynamics is basically a non-reactive case as opposed to a reactive case here and the moment you have reactions then it belongs to only one part which is like the supersonic propagation as we will see and show corresponding to detonation whereas there is also another part which is corresponding to deflagration that subsonic propagation and where expansion happens rather than compression all right let me also say one more thing so now that we have also acclimatized ourselves with some more here we could further actually divide this into five parts I should say four parts rather so you now have you can you can now divide your upper branch as something that is above the UCJ and something that is below the UCJ you can you can divide the lower branch as something that is to the left of LCJ and something to the right of LCJ so let us now call this call this region 1 this is region 2 this is region 3 and this is region 4 so we essentially are looking at the upper branches a detonation branch the lower branches the deflagration branch so what we would like to call this as region 1 at the moment they are just names we just do not still understand them more completely but let us just give the names first and then start understanding because that will kind of aid us aid us in understanding so region 1 is essentially what is called a strong strong detonation and then of course you now have CJ detonation and then we have region 2 what corresponds to your weak detonation and then we have region 3 this corresponds to a weak deflagration and then you have the CJ deflagration and then we have region 4 that should by now be easier for you to figure that should be called strong deflagration for you to remember this at this stage let us just say that when you are now looking at a region that is above the CJ point here the UCJ point here what that simply means is we are now looking at a solution so that means your your your Rayleigh line has to actually pass somewhere there and then you are looking for a intersection somewhere there right that means the Rayleigh line is obviously steeper than the tangent really there that means the wave is propagating faster there right and it is now leading to a very high pressure and a very very high density that is a low 1 over rho infinity and a high P infinity say a very low 1 over rho infinity means a very high rho infinity so this is actually corresponding to a lot of pressurization and lot of compression so that is why we are basically saying strong but it is going to be a little bit more than that we will be able to see what is the propagating Mach number for the wave and what is the downstream Mach number for the products it is based on that that actually we now give this normal nature that this is strong detonation but at the moment you can clearly see that you this is pretty strong in the sense it is having a lot of compression whereas a weak detonation is just as fast if you were to now look at a solution that is here it is Rayleigh line is pretty steep as well just to speak just to steep as well so it is pretty fast all right but it is not leading to a very high pressure and the departure from the origin is not not a lot in the 1 over rho infinity axis so your 1 over rho infinity is not a lot lower therefore the rho infinity is not a lot higher therefore you are having a smaller amount of compression when compared to the strong detonation so relatively speaking we would like to think of this is a weak detonation okay what we will actually find is when you now have a flow that is going through a detonation wave just like in a normal shock you should expect that if the flow is approaching the wave at supersonic speeds if the strong if the shock is very strong you should expect the flow to actually become subsonic all the way that is a mark of its extent of compression so if it has been compressed a lot it also gets diesel the flow also gets decelerated right so a lot of compression would mean a lot of deceleration all the way down to subsonic levels whereas in a weak detonation you expect that the downstream product flow is still remaining supersonic although not as supersonic as the upstream reactants there is the deceleration all right but not down to subsonic levels correspondingly the Cj detonation should mean that of course you are going through a supersonic reactant flow reaching up to the wave but the react the products correspond exactly to sonic conditions that is what is dictated that is what is denoting a Cj the detonation similarly if you now begin to look at this side what it means is in a deflagration we are looking for an expansion so when you now looking for an expansion that means the pressure should decrease and the density should also decrease if you now look at this region the pressure decreases are not a lot and the density decrease the density decrease or the or should say 1 over rho infinity increase is also modest therefore a rho infinity decrease is also modest right so you are now not having a fairly large decrease in P infinity and rho infinity and therefore you are expecting like a weak deflagration okay. So you are now looking at a Rayleigh line that is corresponding to a velocity that is still lower than the Cj deflagration right it is slower and so correspondingly you are not really expecting a large decrease in pressure and density but for this low deflagration you still can intersect over there so this wave is just a slow but it is actually causing a large decrease in pressure and a correspondingly large decrease in density right that is a lot of expansion that is going on and that is why we want to term it as a strong deflagration right. So when you now have a large expansion that is going on proportionately you are now expecting the flow to accelerate so when I react and flow is actually going through a deflagration wave it tends to accelerate and this is actually common practice so in all these things we always try to relate to what we already have experience with so what I would like to suggest is when you are dealing with detonation waves you now recall what happens in gas dynamics what you have learnt in your one-dimensional gas dynamics from your undergraduate let us say if you have gone through a gas dynamics course in your undergraduate level you can recollect what happens across a shock and try to modify your thoughts to accommodate reactions and so on when you are thinking about deflagration waves this is actually more commonplace in terms of daily experience like what happens in like in a Bunsen burner for example right so if you now have a Bunsen burner and then you have like a cold react and flow that is approaching the Bunsen burner and then it goes through the flame and the flame is like a conical flame and the flow goes through this it now expands and then kind of goes like that okay and then you can also see that occasionally and this is not very difficult for you to think about as a thought experiment or even go back and see if you can do this and see for yourself when you now have like a inclined flame like this in a Bunsen burner and if you have like a like a little particle that is kind of glowing okay so we now see how this glow happens as it as it goes along when it passes through the flame it goes much faster when compared to when it tried to approach so it is like a little spark of some particle that is just glowing you can clearly see that it just goes faster right so when you now have a deflagration wave the flow accelerates when it goes past it because it is expanding right now while you are here you are not expanding a whole lot and that is because it is a weak deflagration and therefore you started out with the flow approaching you at subsonic speeds and you are accelerating to further faster but still subsonic product speeds that is the mark of weak deflagration but when you now go to go past the LCJ over here to the strong deflagration what that means is you are now having a very large amount of expansion that is going on correspondingly there must be a very significant amount of acceleration that goes on to the extent that we end up finding that the products actually have supersonic speeds by the time they are getting past this wave where although they started out with subsonic conditions for the reactants correspondingly what just like how we found in the case of UCJ what this means for the LCJ is the approach velocities for the reactants is subsonic all right but the flow accelerates through the wave to exactly sonic conditions beyond for the for the LCJ now if you really think about it this is getting a little bit more fascinating or it is unbelievable fantastic okay how can I have a wave that started out to be subsonic go through a flame and accelerate so much as to become supersonic is it possible for me to do so the answer is we do not come across that at all right so in boons and burners and stuff you do not really get supersonic flows so what is going on the answer there is outside the picture of what we are talking about we started out with the mass conservation equation we started out and then we went through the momentum conservation and then we went through the energy conservation we also talked about the species conservation but the energy conservation keep in mind is essentially coming out of the first law of thermodynamics but we never really considered the second law okay and and we will never do that actually except to state that the this part of the wave that is the strong deflagration is not possible if you now start taking second law into account and as a reason why you will never really find a wave starting from subsonic flows going through a strong deflagration to accelerate all the way to supersonic speeds that typically does not happen because it violates second law right and I think I may be able to show that briefly later on but let me also point out that I will try to show first of all the detonation waves all detonation waves in or in fact I should say all detonation waves travel at supersonic speeds and I should also show that all deflagration waves travel at subsonic speeds the way I would like to show this is to consider the Cj detonation and the Cj deflagration I would like to show that the Cj deflagration is supersonic which means the slowest detonation is supersonic that means all other detonations are going to be supersonic and I am going to show you that the fastest deflagration the C the lower Cj deflagration is going to be subsonic therefore all other deflagrations are subsonic right and then we will also consider the downstream Mach numbers for the ucj and lcj and show that they are actually one thereby we can clearly see how these things are demarcated right but before we do that let us go back and just write out the properties of the Hugonio curve one where they intersect what the asymptotes are and all those things in the next class. Thank you.