 So let us now look at the asymptotes of the Hugoniot, so we could show that the asymptotes in the way we actually try to find out the asymptotes of this Hugoniot curve is to substitute that 1 over rho infinity goes to infinity and then find out what is the value of P infinity for 1 over rho infinity going to infinity to get this asymptote for you to get this asymptote you now substitute P infinity is equal to infinity in the Hugoniot relation and then try to find out what is the value of 1 over rho infinity right, so can be shown to be that means you show as in like you can do this is an exercise and it can be like a typical quiz question or an exam problem 1 over rho infinity equals 1 over rho not times gamma minus 1 divided by gamma plus 1 and P infinity equals minus P not minus P not gamma minus 1 divided by gamma plus 1. Now notice that gamma is ratio of specific heats which is greater than 1, so let us suppose it is about 1.250 something like that, so this is going to be like point something divided by 2 points something, so gamma minus 1 divided by gamma plus 1 is going to be always less than 1 and so 1 over rho infinity is always going to be less than 1 over rho not and therefore your asymptote is going to be to the left of the origin and similarly gamma minus 1 divided by gamma plus 1 again is less than 1 and it is going to be a fraction of P not that means it is going to be less than the less than the value of value at the origin but with a negative sign right, so it is now going to the other side and obviously that is not possible right, so this is less than 0 not possible not physically possible right, so the practical range of P infinity is 0 less than P infinity less than infinity it can go up to infinity that is possible, so where what is the range of rho infinity that is now going to span between where the Higonew intersects, so let us suppose that is how it is intersecting, so then that goes asymptotically to the lower bound there which is impossible but P infinity equal to 0 is about where you can go up to 1 over rho infinity you cannot go below this because P infinity cannot become negative right, therefore this is like your uppermost value of 1 over rho infinity, so you can try to find that out by plugging in P infinity equal to 0 in the Rankine-Hugonew relation right and then try to solve for 1 over rho infinity and get an expression. The lowest value is when P infinity becomes infinity, so that is corresponding to this particular value itself there of the asymptote right, so the range of 1 over rho infinity is 1 over rho 0 ?-1 divided by ? plus 1 which is what we have for the asymptote we saw just a little earlier and you are not going to really touch it because it is an asymptote and therefore you are not going to really touch P infinity equals infinity, so you are always less than that and so 1 over rho infinity is always greater than this not really greater than or equal to but you can go up to greater than or equal to what is now called 2Q divided by P0 plus ?-1 divided by ? plus 1 I am sorry this is ? plus 1 divided by ?-1 1 over rho 0. Now notice that this is actually dependent on Q that means if your Q is larger the Hugonew shifts to the right and top and therefore the point at point where it intersects the P infinity equal to 0 axis shifts further and further out right, so sure enough as you increase your Q this term is going to linearly increase right, so this is a range now what we also noticed was notice that the first quadrant of the Hugonew plot is not possible because Rayleigh line has a negative slope, so this divides the Hugonew into two parts the upper branch and the lower branch, so it can now begin to see what those limits are for each of these branches, so the Hugonew curve is divided into two branches the lower and the upper right, so along the upper branch the ranges the ranges are P0 P0 plus ?-1 Q rho 0 less than or equal to P infinity less than infinity that is going from here that is going from here all the way up to infinity right, so what we are actually locating this is P0 plus ?-1 Q rho 0, so as Q increases this curve is going to go up, so the point where this intersection happens is going to go up right, so P infinity then the lowest value of P infinity will keep moving up as Q increases, so this can obviously be obtained by plugging in 1 over rho infinity is equal to 1 over rho 0 in your Rankine-Hugonew relation right, so each of these expressions can actually be derived by plugging different things in the Rankine-Hugonew relation okay, the question is for what do you need to plug what is something that you have to think about and use your mind and you can get these expressions but these are like little problems that you can face in your exams, this is like a fertile ground for asking lots of little questions right, just to basically just your algebraic skills not really combustion skills okay, then this is as far as the P infinity is concerned and of course the lowest value of rho 0 so you can rho infinity I am sorry 1 over rho infinity that you can get is the asymptote and the highest value is 1 over rho 0 itself that is very easy for you to figure out right, so again we can write 1 over rho 0 gamma-1 divided by gamma-1 less than 1 over rho infinity less than or equal to 1 over rho 0 that is the range for the 1 over rho infinity along the upper branch along the lower branch the ranges are the ranges are again if you are now looking at P infinity it is going to go from here that is the highest value is P0 and the lowest value is this right which is 0 therefore so it is going to go from 0 less than or equal to P infinity less than or equal to P0 okay. And the you now look at the lower branch it starts from here so the lowest value of 1 over rho infinity is corresponding to this value which is obtained by plugging P infinity equal to P0 right so if you now plug P infinity equal to P0 in the rank and you go in your relation you should now get an expression that we should now be able to see readily depends on Q linearly so 1 over rho 0 plus gamma-1 divided by gamma Q divided by P0 less than or equal to 1 over rho infinity less than or equal to that is the other end where the P infinity becomes 0 this is something that we have already figured out that is this value right. So that is 2 Q divided by P0 plus 1 over rho 0 gamma plus 1 divided by gamma-1 good so these are these are just algebraic expressions that you can get for the different limits for the upper branch the lower branch separately okay now as we noticed we can clearly see from here that the upper branch corresponds to a compression wave that actually decelerates the flow and therefore we call it a detonation wave the lower branch is a expansion wave which accelerates the flow and so we call it a deflagration wave okay these are things that we noticed the next thing that we wanted to point out was we wanted to actually find out what is the speed at which the detonation waves travel and what is the speed at which the deflagration waves travel so the way we wanted to actually look at this what is the speed at speed of the lowest or the slowest detonation wave and what is the speed of the fastest deflagration wave and obviously we can see that they are sort of poles apart okay so that is the slowest detonation wave is still going to be much faster than the fastest deflagration wave so let us see what they tell us okay and these are corresponding to the Chapman-Jugey detonation and deflagration waves so what we want to actually look for is what is the incoming Mach number incoming Mach number at the CJ points at the CJ points on both branches alright so this is what we should be looking at now so how do you how do you locate the two the two CJ points right obviously you look at any point any point that is like a solution of both intersection of the Rayleigh line and the Hugoniot is by solving them together right so the Rayleigh line has its P infinity and 1 over rho infinity is showing up and the Hugoniot curve has its 1 over P infinity and 1 over rho infinity is showing up so that is to say that if you now solve for P infinity and 1 over rho infinity together with these two equations you now should get a pair of P infinity and 1 over rho infinity that corresponds to coordinates that satisfy both the equations and that means that that is a point where both of them intersect right but in addition to that at the CJ points we also have to match the slopes because the Rayleigh line is tangential to the to the Hugoniot curve right so so let us first try to get the end states at the CJ points end states as in the infinity conditions right so at the CJ points the slopes of the Rayleigh line and the Hugoniot curve match so that means we had a first of all find out what is the slope at any point on the Hugoniot curve and then match it to the slope of the Rayleigh line so that the slope of the Hugoniot curve is DP infinity D over D 1 over rho infinity this is something that you can actually obtain by simply differentiating the Hugoniot curve with respect to 1 over rho infinity right so wherever you are getting a P infinity you try to now differentiate that with respect to 1 over rho infinity right so then you get a P infinity minus P not times sorry minus 2 gamma over gamma minus 1 times P infinity divided by 2 gamma divided by gamma minus 1 1 over rho infinity minus 1 over rho 0 plus 1 over rho infinity that is the slope expression that you get for any point on the Hugoniot curve this is actually the slope at any point okay we are still not saying that this is at the at the CJ points at the CJ points it turns out that the slope should now be equal to the Rayleigh slope right so slope of the Rayleigh line what is the slope of the Rayleigh line Rayleigh line is a straight line okay and so long as we do not want to use the m dot squared expression that is like a no no yeah because we want to deal with only the thermodynamic properties so for a straight line it is just P infinity minus P not divided by 1 over rho infinity minus 1 over rho not that is it right so is P infinity minus P not divided by 1 over rho infinity minus 1 over rho not so equating the two equating the two we can we can we can get so we can get like a for example P infinity P infinity equal to P not rho not divided by rho infinity times gamma plus 1 rho not divided by rho infinity minus gamma so here what we have done is we have eliminated or we have now written expression for P infinity in terms of the other three P not rho not and rho infinity okay in fact we always should be looking for rho infinity to show up as 1 over rho infinity so sure enough this is 1 over rho infinity here this is 1 over rho infinity here and we are also getting it as rho not over rho infinity everywhere right here as well as here okay so essentially that means we should now be in a position to eliminate P infinity in favor of rho infinity or rather 1 over rho infinity or if you want to do it the other way you can now try to get an expression for 1 over rho infinity in terms of P infinity P not and 1 over rho not okay so or 1 over rho infinity equal to 1 over rho not times gamma P infinity divided by P not times 1 plus gamma rather you used to writing gamma plus 1 P infinity divided by P not minus 1 okay so you can use either of these expressions so these are both at CJ points both is both these expressions are valid at CJ points okay doesn't really solve much we have not really got the coordinates yet because here P infinity depends on rho infinity so what is it meant by saying getting the coordinates we should now be able to write P infinity only as a function of P not rho not and Q okay and 1 over rho infinity we should be able to write in terms of P not 1 over rho not and Q but this is the mixed thing we just can eliminate because we give you or we can write explicit expressions for P not at the seed P infinity at the CJ point and 1 over rho infinity at the CJ point in terms of the others which will still contain the other infinity term P infinity will contain 1 over rho infinity term and P 1 over rho infinity will contain P infinity okay so there are these other things so we still haven't solve the solve for the coordinates which is what we are now going to do next so substitute either of these right in the Hugoniot or strictly speaking say Rankine Hugoniot to give Rankine his new relation so that what happens is if you now write P infinity in terms of 1 over rho infinity in the Rankine Hugoniot you don't have P infinity at all in that expression you will now have an expression that is only on 1 over rho infinity which means you can now write what is 1 over rho infinity in terms of everything else right so to so but unfortunately you are now going to get a quadratic okay in either of these but it is actually fortunate if you physically fortunate mathematically a bit unfortunate you are dealing with quadratic equation but it is physically fortunate because this says only CJ matching the slope simply means that we have we are just working at any CJ point it could be the upper CJ point or the lower CG point either of these CJ points the really line is tangential to the Hugoniot it is not distinguishing between those those two right and we are therefore looking for two solutions to 1 over rho infinity and P infinity at the CJ points one of which will correspond to the upper CJ the other one corresponding to the lower CG point so here so substitute either of these expressions in the in the in the Rankine Hugoniot relation to get get a quadratic quadratic in either P infinity or 1 over rho infinity and solve solve the quadratic so if you now solve the quadratic you get P infinity plus or minus equal to P not plus gamma minus 1 Q rho not times 1 plus or minus 1 plus 2 gamma P not divided by Q rho not gamma squared minus 1 the whole to the half okay now let us call this small a the expression for P infinity plus or minus 1 and 1 over rho infinity plus or minus equal to 1 over rho not plus gamma minus 1 divided by gamma Q divided by P not times 1 minus or plus 1 plus 2 gamma P not divided by Q rho not gamma squared minus 1 so that is kind of like the same as what we had before to the half right okay so here what happens is when you now say plus or minus this plus goes with the plus here this minus goes with the minus here right but here this plus or minus is such that this plus goes with the minus there and this minus goes with the plus here right so when you say basically say plus or minus you simply mean upper CJ or lower CJ that means we are expecting that for the upper CJ your P infinity should always be greater than P not therefore P infinity plus will correspond to P not plus this okay lower CJ is going to be having a P infinity less than P not therefore P infinity minus should be equal to P not plus this minus with the negative sign that means you are going to have a less than and similarly here 1 over rho infinity you are not going to have 1 over rho infinity that is for the for the upper CJ you need to have a 1 over rho infinity that is less right therefore you need to have a negative sign over there and and for the deflagration CJ point the low LCJ point 1 over rho infinity is greater therefore you need to have a plus so the lower sign will always correspond to a LCJ upper upper sign corresponds to a UCJ right so upper sign corresponds to UCJ lower sign corresponds to LCJ so we now got the coordinates great so of course we need to call this be good I mean the in fact getting this with this particular thing or substitute those two expressions in the you go in your curve expression the rank and you go in your relation and get the quadratic and solve the quadratic to get these this is actually done for each of those separately right you will sweat a little bit it is going to take like some amount of time doing all this stuff okay well it could be a bit boring but you wait through the mathematics to see how the terms basically get grouped the way they are okay and then it is going to be pretty instructive for us because of what we are going to do next now what we want to do is what if what do we want to do what we want to do we want to now look at what is incoming Mach number why do you want to look at the incoming Mach number because we want to know how fast the CJ detonation wave and the CJ deflagration wave are going to travel right now keep in mind how fast as the wave going to travel is an information that is contained in U0 okay because in a wave fixed coordinate system you are going to have the reactance travel to the wave at that speed and that information is now in our scheme of things embedded in m dot because m dot is the one that is containing rho 0 U0 as well as rho infinity U infinity so it serves purposes of take carrying the flow information on either side of the wave right and since we are actually looking at the incoming Mach number we should be focusing on U0 that means we are looking at m dot through which we will try to look for rho 0 U0 and m dot does not show up in the Hugonio curve it is showing up only in the Rayleigh line right so we should now go back to the Rayleigh line and say it is sort of like saying look at whatever we have done we have substituted the slope matched information into the Hugonio curve to get this but can I now look at how the Rayleigh line is going to be like if I plug these coordinate points in the Rayleigh line and then get an m dot information through that is you what I am saying okay so in the we have really not used the Rayleigh line information except to note that that is the slope but that is just a thermodynamic that is just a slope written in terms of thermodynamic variables we have not brought in the fact that this is equal to minus m dot square right so that is what we want to do now so we want to now notice that the slope that we had that we had used for the Rayleigh line is actually equal to minus m dot square and then get to the flow information so using A and B in the in okay so simply just go back and write what it is P infinity minus P not divided by 1 over rho infinity minus 1 over rho dot equals minus m dot square right we can now get our m dot square divided by P not rho not plus or minus equal to gamma plus Q rho not divided by P not times gamma square minus 1 times 1 plus or minus 2 gamma P not divided by Q rho not gamma square minus 1 the whole to the half okay note that note that m dot equals rho not u not and the Mach number m not is going to be written as u not divided by a not which is nothing but u not divided by square root of gamma P not divided by rho not if you now notice that m dot is nothing but rho not u not so now now you write your u not is m dot divided by rho not okay so you now plug that over here you will now get your this is nothing but square root of m dot square divided by gamma P not rho not so we had m dot square by P not rho not all right okay and then we now are looking for m dot square by gamma P not rho not under square root is your Mach number or m not squared is now going to be m dot square divided by gamma P not rho that is what I mean this is something that is basic it is nothing particular about whatever we are doing this is like just putting these things together right so from here we should now be able to write m not plus or minus as 1 plus Q rho not divided by P not gamma squared minus 1 divided by 2 gamma the whole to the half plus or minus Q rho not divided by P not gamma squared minus 1 by 2 gamma the whole to the half so it is sort of like falling into some sort of a pattern there and this is what we were looking for so because it is going to say something to us so at the Cj points okay so what we can see from here is when Q goes to 0 right so when Q goes to 0 this goes to 0 this goes to 0 right so you just get m not plus or minus 1 is equal to 1 all right so when Q goes to 0 we get m not plus or minus 1 equal to 1 I can say goes to 1 and we will also see that m infinity plus or minus 1 at Cj points equal to 1 that is something that we will do next okay so what is about that simply means is nothing is happening when you do not have any heat release it is as if like you had a wave that is going at the sonic speed okay and not really changing anything about your gases the gases are basically non-reactive gases that means you do not you are not having any heat release and that is as if like you had gases go at sonic speed and nothing happened to it okay that is not a very interesting situation for us but let us now look at the other possibility what we are really interested in is when you have non-zero Q now how would you actually look for a non-zero Q in the Rayleigh line as I said when you now have a Q greater than 0 as Q increases you now have this Rayleigh line push up like that away and away from the origin of the Egonium right and as you now go up to Q turning to infinity let us say you now had Q going to infinity right this curve is going to actually go and hit the Rayleigh line that the Cj point is go get pushed to infinity and this Cj point is going to get pushed to infinity here alright and what you are going to now see is either your 1 over rho your P infinity is going to go to infinity or your 1 over rho infinity is going to go to infinity the P0 is going to go to 0 over here your 1 over rho your P infinity is going to go to 0 here and your 1 over rho infinity is going to go to the asymptote this is this is how it is going to push right so if you now look at the limit then so as strictly speaking do not have to look at Q going to infinity itself what really matters for us is in this expression Q ?0 divided by P0 goes to infinity then if you now use the positive sign for the UCJ that is a detonation wave then both these terms add up together towards each other and then the M0 plus goes to infinity right. So M0 plus goes to infinity but M0- goes to a 0 these two terms essentially subtract from each other of course one is like a small number when compared to these infinities that we are talking about so M0- goes to 0 right. So what this means is what you are then saying is then the for the Chapman-Jugey detonation wave detonation wave keep in mind when Q is equal to a 0 M0 plus or minus both of them are going to be equal to 1 that is a lower limit as far as the heat is concerned right. So for the Chapman-Jugey detonation wave we have M0 plus go from 1 to infinity right and for the Chapman-Jugey deflagration wave we have 0 less than M0- less than 1 you see so these are the limits that we are talking about therefore this is how it is going to pan and what that means is the M0 is not really going to overlap between these two okay. So this is going to go only up to 1 that is going to go only above 1 okay so that means a detonation wave travels at propagates at supersonic speeds and a deflagration wave propagates at subsonic speeds well let us not just then jump to that conclusion yet there is just one more step that we have already talked about the reason why and then I want to bring that in to generalize this quite well the reason why we were actually looking at M0 plus or minus at CJ points is keep in mind the CJ detonation is the slowest of all the detonation waves and the CJ deflagration is the slowest of all deflagration waves right so if you now have a CJ wave that is supersonic right that is a CJ detonation wave that is supersonic all other detonations are going to be supersonic because this is the slowest and if a CJ deflagration wave is going to be subsonic then all deflagrations are going to be subsonic right so the this implies that right all detonation waves propagate at supersonic speeds and all deflagration waves propagate at subsonic speeds so that is exactly what we were expecting or thinking will happen we also thought about a few things which is what happens downstream over here so if this wave is going to travel at supersonic speeds for detonation waves and compresses the gas the reactants as they now become products they are compressed they also decelerate so the question is or do they decelerate down to what okay so the answer there is they could decelerate down all the way down to subsonic conditions or they could get decelerated to supersonic conditions that are still slower than the incoming supersonic speed this is typically what you are expecting when you are now looking for a detonation wave that is not a CJ detonation so if you now have a any detonation right you can now have solution here and here and this is something that is going to correspond to a very high pressure and very low density sorry very high density as well very low one over density therefore we are expecting it to get a lot compressed and therefore lot decelerated so it can get down to subsonic speeds over here and on this side it is not going to be that much compressed therefore it is not that much decelerated so you could expect that this is not decelerated down to subsonic speeds but in supersonic yet less than slower than this the incoming supersonic speed and therefore we need to anticipate that at the CJ point the outgoing wave is actually going at sonic conditions right so this is what we need to anticipate and let us see if we can if that is going to work out so downstream Mach numbers at the CJ points right so what we then get is we had this expression that we we had this expression by matching the slopes of the Rayleigh and the Hugonio curves right for either the P infinity or the 1 over rho infinity which we substituted earlier in the Hugonio curve to get the coordinates of the CJ points all right and from there we now equated plug that value in the Rayleigh to get the m dot and from there we now went to the upstream Mach number but we can do something different here we have at the CJ points we have at the CJ points P infinity equal to P not divided by gamma plus 1 minus gamma rho infinity divided by rho not or 1 over rho infinity equal to 1 over rho not times gamma divided by gamma plus 1 minus P not divided by P infinity I think this is slightly different from what I just said a minute ago but it is possible for you to derive this by by dividing the numerator and the denominator appropriately by by by let us say P infinity and 1 over rho infinity and so on so it is supposed to have these what you did was you actually plug this into the Hugonio curve expression and got the quadratics and solve them to get the coordinates for the LCJ and the UCJ but if you can now go back and actually plug these in the Rayleigh line itself right so substitute either of these in the Rayleigh line equation and and note that this time we are interested in noting that m dot is not rho not u not because we are interested in the downstream conditions so m dot is actually rho infinity u infinity so note that m dot equals rho infinity u infinity right you should get right so to get simply get m dot squared equals gamma P infinity rho infinity now this is not very different from what we had over here you see you got as I told you we could have written this as m not m infinity is equal to u infinity divided by a infinity equals u infinity divided by square root of gamma P infinity rho rho infinity and then notice that plug this back in this here and then you can get m square m dot m dot square divided by gamma P infinity rho infinity if you got that and then by substituting this expression you also got this that simply means that m infinity is equal to 1 right so this is simply going to mean that m infinity plus or minus equal to 1 now in a mathematical sense this is not surprising because we plugged it we plug this expression in a Rayleigh line which is essentially linear right that is just a straight line therefore we are not expecting any quadratics right so if you do not have a quadratic then we cannot get two solutions we should get only one solution and that is only one solution okay we strictly speaking should not even write a plus or minus we are not even at this stage of distinguishing between plus and minus both of them are the same so that means we are basically looking for the same answer for the downstream Mach number in either of these points right there is to say the downstream Mach number at both CJ points right at both CJ points is unity this implies the CJ detonation decelerates supersonic reactant flow to sonic product flow and CJ deflagration accelerates subsonic reactant flow to supersonic product flow I am just going to abbreviate there because it is pretty obvious now if that is going to be the case so the next step that we do is to then say that if you got a strong detonation then strong detonation means supersonic reactant flow becomes subsonic product flow right CJ detonation is what we just saw supersonic reactant flow becomes sonic product flow and weak detonation becomes supersonic reactant flow all right but still supersonic product flow with M infinity less than M not okay you still have a deceleration weak deflagration means subsonic reactant flow becomes subsonic reactant flow still becomes subsonic product flow with M infinity greater than M not still less than 1 right so we should strictly speaking say with 1 less than M infinity less than M not the previous case here we should say with M not less than M infinity less than 1 right what we should say CJ detonation deflagration we saw subsonic reactant flow become sonic reactant flow sorry product flow and strong deflagration subsonic product reaction flow becomes supersonic product flow which is impossible because it violates second law which we have not explicitly considered mathematically and we will not do that just to accept to state that this is not going to be possible physically we will stop here for the day and we will try to wrap up this one on Monday.