 Let us now look at what happens to the gases that are flowing downstream of the combustion wave if the wave happens to be a deflagration wave or a detonation wave what we have seen so far is if you have a deflagration wave it is an expansion wave and so you have the reactants react into products and the product gases downstream of the wave will have the pressure decrease a little bit and density decrease and they would get accelerated and if the acceleration is just only mild then the deflagration wave of course propagates at a subsonic speed relative to still reactants and the products will also be accelerating only up to subsonic speeds but if it is a Chapman Juge deflagration it can get accelerated all the way up to sonic velocity or if it is a strong deflagration it can get accelerated all the way up to supersonic velocities which of course we have been saying is forbidden by the second law on the other hand if you now look at the upper branch of the Hugonio curve we have a detonation wave that propagates at supersonic speeds relative to still reactants and the products go through a compression and that means they are having a pressure rise and the density rise and then they get decelerated and for a weak detonation they do not get decelerated significantly so the product velocities still turn out to be supersonic whereas for Chapman Juge detonation it is turning out to be sonic and for strong detonation gets decelerated all the way down to subsonic speeds this is what we have noticed what we want to now look at is what is the direction of this velocity of the downstream gases what does that mean right so we have this wave that we wanted to keep and then we have been thinking about a wave fixed coordinate system in which the wave is like this and then the gases are coming in the reactants are coming in and the products are moving so we thought the deductions are all fixed right alright but then keep in mind that if you now allow for a lab fixed coordinate system in which you have still reactants and then you want to have the wave move right that means you want to now subtract out the incoming velocity from the entire flow field and now start looking at whether the reactants are going to still be going like this or going like this in other words sorry products the product so reactants is still and the wave is moving and we now subtract the reactant velocity from the product velocity as well and then we want to find out whether the products are going to be going like this or going like that or still in other words we want to know whether the products are going to actually follow the wave or go away from the wave right so that is what we want to do now and let us see how to do this so downstream velocity at non-CJ points or whatever we do actually also applies to see the CJ points as well it is just that when we say non-CJ points it is lot more general when compared to what we have done before which is to look at the Mach number downstream Mach number at the CJ points from where we just generalize saying that weak detonations will have supersonic downstream velocities strong detonations will have subsonic downstream velocities and so forth by just noting that the CJ detonation or deflagration will have sonic downstream Mach numbers but here what we are trying to do is to determine the direction of this velocity relative to the wave for any point any point not just a CJ point so this is applicable to non-CJ points as well so let us now look at the detonation branch so in the detonation branch so in the detonation branch of the Hegoniou curve right we have 1 over rho infinity is less than 1 over rho 0 right therefore we have U infinity minus U0 equals m dot times 1 over rho infinity minus 1 over rho 0 you know how to get this which is less than 0 which which implies that U infinity is less than U0 this is this is no news we know that the downstream velocity is going to be less than the upstream velocity because the flow is getting decelerated but that is in a lab fixed core sorry a flame fixed coordinate system or a wave fixed coordinate system so if you now have a wave that is fixed in a coordinate system and you have had a U0 like this and U infinity like this this is what is the story right but if you now look at this so this is very fixed right but if you now have something like a lab fixed coordinate system where you are you are saying let us now have the wave move with the velocity v wave okay and we now suppose that your reactant velocity is in this coordinate system designated as v0 and suppose let us suppose let us say it is set to 0 that means we are having still reactants right and in our notation let us now say that we want to have our v infinity follow the wave right that is simply saying I want to subtract U0 from this and if I were to now subtract U0 from here it is going to become 0 that is what I am going to get here if I now subtract U0 from here is it the question is it the is the arrow going to flip or not okay is it going to continue to be this way or not okay that means we want to basically find out whether v infinity – U0 is going to be positive or negative that is exactly what we are looking for okay assuming though that this v infinity is going to be in this direction so let us now do the coordinate transformation so v0 is equal to v wave – U0 and v infinity equals v wave – U infinity right so essentially now the v wave is now thrown upon this picture therefore whatever we are actually now beginning to see here for the velocity is going to be v wave subtracted velocities therefore now if you have still reactants so this is the lab fixed coordinate system for still reactants v0 equal to 0 just in case you when we have been wondering how come we have a v0 yes we will try to subs we will now try to set it to 0 to indicate that we have still reactants which means that our v wave is equal to U0 now notice this carefully we can say v wave is equal to U0 because of the way the arrows are marked if you now take your arrow for the v wave to be right word that is when it is equal to U0 with the U0 actually marked this way all right that is basically say that the v wave exactly opposes U0 in the wave fixed coordinate system therefore your v wave is becoming stationary right so this is having the directions taken for these two things embedded in it when you now say v wave is equal to all right. If you now say that then then we infinity we infinity we can now plug we infinity is taking the direction of we infinity to be following the wave if you now do that then we wave can be substituted as U0 and you can now get U0- U infinity right but what did we what did we notice for U0- U infinity U U0 is greater than U infinity which is now greater than 0 right so U0- U infinity is greater than 0. So if you now have U0- U infinity greater than 0 for v infinity if v infinity is greater than 0 with the direction taken this way then that means this arrow arrow sign is correct right so you have a positive quantity for arrow sign going towards the wave which simply means that the burn gases behind a detonation wave will follow the detonation wave right. So this implies that the burn burn gases burn gases did behind a detonation wave follow the wave right but what but but when you now try to say the way the burn gases are going to follow the wave will they catch up with the wave will they go past the wave that does not make sense at least if they were to be able to go past the wave then they will become reactants but they are supposed to be products so we better want them to actually kind of follow the wave rather faithfully right but what is going to happen in reality. So the question is then if you now say v wave is equal to so what do we have for the v wave v wave is equal to U infinity plus v infinity right so if you now use this one here you can find that v wave is basically v infinity plus U infinity and we now find that both you both your U infinity and v infinity the way they are defined in these two coordinate systems both of them are positive right. So this implies that v wave will obviously be greater than v infinity and how did we define the v wave v wave and the v infinity are in the same direction right since v wave is greater than v infinity it follows that the burn gases will never be able to catch up with the wave the wave is going to travel faster than the burn gases right. So this implies that the burn gases although traveling in the same direction as the wave can never catch up with the wave this is true even when you now have a weak detonation where the product gases are traveling at supersonic speeds they travel at supersonic speeds that are still less than the supersonic speed at which the wave travels okay. Now when I say that you got to be a little bit careful because when you say supersonic or subsonic it is it is based on the Mach number and the Mach number is the way we are actually trying to deal with for the upstream Mach number and the downstream Mach number are relative to their respective temperatures and in all these cases the upstream temperature is lower than the downstream temperature okay. So the downstream Mach number always would look like it is it can be lower okay it does not mean that the velocities are lower but what we are actually finding out in this is the velocity is lower than the wave velocity okay therefore it is going to it is not going to be able to catch up it follows but it cannot catch up all right what happens when you now have a deflagration. So in the deflagration branch of the Hugoniot of the Hugoniot curve we have 1 over rho infinity minus 1 over rho not right so where is the 1 over rho infinity 1 over rho infinity is on the right side when you get the solution for the lower branch when compared to 1 over rho not so that is actually greater than 0 therefore if you now have 1 over rho infinity minus 1 over rho not is greater than 0 then U infinity minus U not equals m dot 1 over rho infinity minus 1 over rho not is greater than 0 right. So U infinity is greater than U not that is no news in the wave fixed coordinate system we know that the flow the product flow is getting accelerated when compared to the reactants but now you look look at what the consequence is when you now have the lab fixed coordinate system if you now say U infinity is greater than U not and then go back here and look at what happens to be infinity right so V infinity is equal to U not minus U infinity that is coming out of the coordinate transformation regardless of the actual disparity between U infinity and V U not okay and so all this is for a general wave this is for a general wave with still reactants this is for a general wave we have not said that it is a detonation at that point it is only when you say greater than 0 we have invoked this that is for a detonation but it turns out to be this for a deflagration if you now plug it in here you find that then V infinity equal to U not minus U infinity what do you get U infinity minus minus U sorry U infinity is greater than U not therefore U not minus U infinity is now going to be less than 0 that means V infinity is negative V infinity being negative the way the arrow is taken this way means that it is going to the arrow is going to get flipped right so this implies that the burn gases the burn gases behind the deflagration wave flow away from the wave away from the wave in the opposite direction so since the gases are going to go away from the wave we do not have to worry about what their speed is relative to the wave speed because they are never going to get never going to catch up right. So let us not even do the extra thing that we did for the detonation so you do not have to worry about whether they are going to catch up or not they simply go away so that means in reality what happens is if you now have a deflagration that is propagating this way in still reactants as soon as the reactants react you now have gases flow away in this direction if the wave is going like this simply because they are getting accelerated and you had still reactants and you now have a pressure decrease and a density decrease so it is an expansion that is going on with an acceleration so the velocity that is accrued because of this tends to flow away from this region right that is what you get for a deflagration now we will see a couple of other thermodynamic properties with yes no this is this is basically based on the way we have actually accounted for these directions for the wave and the in fact what we should what I should say is we are also assuming a v0 like this that is how you can get this coordinate transformation going but of course we set v0 equal to 0 because we want to consider still reactants from where we get v wave is equal to u0 but I wanted to point out previously that v wave is assumed to be like this and u0 is assumed to be like this that is how we got this transformation to be like this and yes correct so when I put the relative velocity of the products it would become vw plus true then what happens is see why would you want to do that so you would want to do if you want to now factor in the relative velocity of the products relative to the wave then you are going back to the wave fixed coordinate system right if you what is meant by relative velocity relative to what relative to what yeah relative to the wave so if you now want to actually locate your velocities relative to the wave you are going to have a wave fixed coordinate system unless unless you are now trying to do well maybe maybe not maybe maybe what you are saying is different from what I am saying because when I when I am in a wave fixed coordinate system I am actually counting my reactant velocity relative to the wave okay when I am in a lap fixed coordinate system I am just letting the wave propagate in it primarily still reactants but it does not have to be still okay but if I am counting for a flow direction I would count it like this if the flow were actually approaching the wave like this in a in a wave in a lap fixed coordinate system I would have to reckon V0 to be negative in the way I am doing this what you are telling me is once I actually find out what my actual V infinity is is it possible for me to look at that velocity relative to the wave that means we are we are now trying to look for making this still if that is if that is if that is what I understand the equation what we have taken is for V infinity is Vw-U infinity the infinity is V what the infinity equation is Vw-U infinity okay that is because I am taking the direction like this yes of V infinity but if I take the direction itself like this then the relative velocity would become Vw plus U infinity so the equation there is such a change so this process okay if you were to take your V infinity direction to be like this you should get V infinity to be positive that that that is what you should expect all right which would which would be basically say the same thing all right so I mean you you can work that out maybe what you what you do is you assume that V0 is going to be positive in this direction V infinity is going to be positive in that direction you should be able to show that if you were to actually make V0 equal to 0 right then V infinity becomes negative for detonation wave and it is it is positive for reflagration wave that will that will simply make this different you would you would now write to you would now say that V0 is equal to U infinity-Vw and V infinity is equal to U infinity-Vw all right that is how you will actually get these signs you should get the same thing right the question is all about being consistent about your about these transformations based on the direction that you have assumed okay so I guess I guess that would be a little bit more straightforward maybe if you would assume the wave to go like this and still assume the reactants to go like this and the products to go like that it is it is probably a lot straightforward to do the transformation right and all you have to do then is to show that V infinity is negative in the case of detonations and remains positive in the case of deflagrations so that you have to show the opposite okay maybe for the exam right so yeah but it should be the same all right let us just go through a couple of quick thermodynamic properties other than what we have discussed so far before I come back to some some realistic realistic depictions of these waves and see what happens so first thing I would like to point out is temperature temperature behind the waves right T infinity is greater than T not always for Q greater than 0 what you can show is show if the main molecular weights of the reactants and products are assumed to be equal that is 1 over sigma i equals 1 to n yi infinity divided by wi equals 1 over sigma i equals 1 to n yi not divided by wi equal to W bar then then the temperature ratio the temperature ratio T infinity by T not equals P infinity by rho infinity divided by P not divided by rho not is greater than 1 for at Cj points that is to say you can show this at Cj points considering the simplifying assumption that the mean molecular weight of the reactant and product gases are equal but that is just a mathematical exercise that you can go through the second thing that I would like to point out is how the entropy varies there is something that we have not talked about at all but it is interesting to talk about entropy variation entropy variation along the Hugoniot so maybe we have these underline to distinguish them. So it can be shown that and it is not very difficult to show I will show I will tell you how to show this it can be shown that the entropy is a minimum at u Cj u Cj and maximum at L Cj so that is basically again looking at the Cj points because they are more or minimal to mathematical manipulation given that you can match the slopes of the Rayleigh and the Hugoniot and get coordinates and so on and then how do you show this okay so I am not going to show this but you can show question is how do you show this well it is rather straightforward what you have to do is you see that the Hugoniot curve is having P infinity on the vertical axis and 1 over o infinity on the horizontal axis so it is as if 1 over o infinity is your independent variable and P infinity is your dependent variable right. So what you would what is suggested is you now try to express your entropy as a function of 1 over o infinity as a independent variable still and if you do that and then of course the way you do this is you now assume a reversible hypothetical reversible reaction that can prevail between the end states and then use like TD T infinity DS infinity equal to in the first law of thermodynamics and so if you now do that then you get a expression for entropy at the end state as a function of 1 over o infinity and use that expression to differentiate this entropy with respect to 1 over o infinity at the Cj point note that I am not saying which Cj point because whether it is maximum or minimum you are going to have the derivative equal to 0 so you should be able to show this this is relatively easier when compared to the next step which is you have to now show whether it is maximum or minimum right. So what you will be able to show is with D squared S infinity divided by D 1 over o infinity squared at u Cj as greater than 0 and D squared S infinity divided by D 1 over o infinity squared as less at LCJ as less than 0 right. So what is happening what you will find this I am just running out of space there let me use this panel to depict what is going on so if you now think about the way I said that is you have your 1 over o infinity as your independent variable and you plot your S infinity as a dependent variable question is how does it look like okay now this is good to do because the 1 over o infinity is going to be the same in the Hugonio as well as here and we know what happens when you now have the 1 over o infinity vary about its origin the origin of the Hugonio of course is 1 over o not right and then the 1 over o infinity varies on either side 1 over o infinity increases in the deflagration the lower branch and 1 over infinity decreases in the upper branch for the detonation but keep in mind that there is a there is a region here which belongs to the first quadrant in the Hugonio which is not covered right that means 1 over o infinity goes from 1 over o not as the initial condition to a value it jumps to a value that is where the Hugonio cuts the x axis passing through the origin right because about this you have the first quadrant in the Hugonio plot which is forbidden and therefore you will you will you will you will find that you now have you will now have a curve that goes like that and this is where your LCJ is and then you have over here a curve that looks like this which is where you now have your UCJ right so what we are basically saying is this is your region 1 this is region 2 this is region 3 this is region 4 so that that means this is strong detonate detonation you have CJ CJ detonation this is weak detonation we have weak deflagration and of course we have CJ deflagration then we have a strong deflagration right then they let then let us think about what really happens in actual practice okay what we have seen so far is something kind of trivial we have assumed that we can treat whatever happens inside the wave as a blood box and then just do balances across as if they just jump but we need still to conserve mass momentum energy and so forth and we have now come upon a lot of facts like we should now be able to say that there are two kinds of waves that are completely apart one goes as deflagration subsonic and another one is def detonation and so on and then we further looked at upstream and downstream conditions to point out that this is subsonic that supersonic then you have strong weak and all those things right but did we ever care about whether this really happens in reality okay so realistic existence of these waves of the different waves the first thing is strong detonation what you will find is strong detonations are not very easily observed okay so seldom seldom observed and why so because it is a strong detonation that means you have to have a large compression and you have to make sure that the gases are decelerated down to subsonic speeds from a supersonically propagating wave and you have to have large pressures there and maintain okay so invariably the way detonations typically happen is if you now have a reactant gas in a confined region and you and they got ignited at the confined end and you now have a wave that is beginning to propagate you now have a confined region where the pressure can build up and stay right and once you can have this pressure build up then you have a transition to detonation wave for what you what you what you are igniting you now have a detonation wave but the moment the detonation wave begins to propagate this this space is vacated right so you cannot really have the pressure hold on until you are able to actually have this confinement follow the wave right. So this is what is called as a over driven shock tube so that means you have to have like a piston that runs along with the wave and keeps this confinement as it is in order to be able to preserve the very high pressure that is been obtained with the strong detonation so it requires a special experimental arrangement requires arrangement what is called as over driving the shock now Cj detonation on the other hand is always observed that means whenever you have a detonation wave there is there is a very high chance that it is always a Cj detonation so always observed in fact in the in the experiment that I just discussed if you do not have this over driving you have this detonation wave right and you start out with a strong detonation let us say because we had the confinement and keep in mind what happens with the strong detonation wave you now have a subsonic flow be downstream right and you now have the pressure pulses that go and get reflected from this and begin to come back and hit the wave because this is a subsonic flow and the pressure pulses can actually propagate faster than the product flow product is actually beginning to follow whether subsonic speeds and it is not going to catch up but it can allow for the waves to go this pressure disturbances that are coming from the other side can actually pass through this because this is subsonic typically what then happens is you now have an expansion fan should say fan if you should say expansion wave because we are basically looking one dimensional so you now have this wave that go like this and then there is a pressure build up and that also propagates so you have an expansion region that is getting created which can propagate through this and then go and try to weaken the wave all right when that happens progressively this wave is getting weakened down to CJ level at which point the products are beginning to actually go at sonic speed right so once the products are going to going at sonic speed the data faction wave that we talked about can never really go past them right because they can go only at the speed of the products and not really reach the wave at all and therefore you now stop at a CJ so any strong detonation that there has been created will always tend towards the CJ detonation pretty soon and there is a reason why you will always find a CJ detonation unless you took this arrangement in order to preserve the strong detonation right so if you did not do any such arrangement you are going to be getting a CJ detonation third weak detonation now this is also seldom observed here what happens is you are having the react the product gases travel at supersonic speeds while it is while it is trying to follow the wave and if you want to have supersonic speeds at which the products are traveling your you have to have chemical reactions that are happening in the wave happens so fast that they are able to actually produce products that can travel so fast so this really means that you have to have chemical reactions that happen very fast so you need to have very special mixtures that have extremely fast kinetics so require special reactant mixer mixtures that have extremely fast kinetics fast when compared to the flow velocities of the products so that you will get products otherwise you will still have reactions happening while you are still expecting products to happen the products are not happening because the reactions are taking slow but to taking time to happen so that is about the detonations let us now see what happens to the deflagrations should talk about weak deflagration sorry it is not really stopped by anything weak what that is only for the react that is that is only for the infinities infinity condition but 1 by 0 and not I have S0 and whatever I get in this plot is below the value of true that is correct so it should not happen at all no that is not true entropy is decreasing only for the system because the system is being worked upon when you have a compression wave what you really want to be looking for is a the entropy for what is actually causing the work to be done as well so it is the entropy of the universe that actually ultimately increases it does not have it does not mean that if you have a particular system entropy can decrease if you do work on a system then entropy can decrease but who is doing the work on the system will have a corresponding entropy increase which will be greater than the entropy decrease that you find in the system because it is a compression you will have a entropy decrease because work is being done on the system all right so that is for another day so according to what you say none of this can happen right well you have to start from here this is your starting point if you were somewhere here you are still low right that is not true so talking about weak weak deflagration always observed here in fact if you now go back and look at your Hugonio this is the origin and let us say that is a CJ point what you are talking about is somewhere that is like very very close to the original value that means if you have this is your P0 and this is your P infinity P infinity is just about little less than P0. So most of your weak deflagration is happening very close to the original P0 and very far away from the LCJ in other words you do not even go all the way somewhere here you are very close there that is mostly what happens you are typically talking about open systems and in open systems you have almost like a constant pressure process it is only very close to the flame it is possible to have this acceleration and this expansion therefore the pressure decreases a little bit not a whole lot right and that is exactly what gives you a weak deflagration correspondingly if you are now thinking about a CJ deflagration you simply have to say not observed this is primarily coming from the numerics that means if you just put the numbers in you will find that you cannot get the pressure to decrease significantly at all for the downstream case to be able to get down to a CJ level right and when you now finally talk about a strong deflagration all you have to say is not possible forbidden by second law this is something that we have been talking about so in reality in the CJ in the Higoneo line we are pretty much confined to a region that is very close to where we started for the P0 and you do not even go all the way to the CJ cannot really go past it at all in reality okay so most of your deflagrations are extremely weak deflagrations that is the way I would put it so let us finally give some numbers on what these changes are like so if you now look at detonation on the one hand deflagration on the other hand whatever M0 it is of the order of about 5 to 10 here it is of the order 0.00012 about 0.03 there are about 2 orders magnitude variation but they are very very small u infinity over u0 how does the deceleration or the acceleration happen this is about 0.4 to 0.7 deceleration here it is about 4 to 6 you can work this out if you assume constant pressure you can work this out to be about 4 to 6 base based on the temperature change this is acceleration P infinity over P0 this is about look at this number 13 to 55 compression this is about 0.98 there is about 98% pressure recovery there so that is slight expansion and this high pressure jump across the detonation wave is what causes all the damage in detonations okay T infinity over T0 is about 8 to 21 that is heat addition it is only slightly less over here it is about 4 to 16 also heat addition in both cases we are actually adding heat and finally look at rho infinity over rho0 you have about 1.7 to 2.6 and over here it is about 0.06 to about 0.25 that is about it so get it you get the picture on what do you what do you get for these detonation or deflagration waves in summary you typically are looking at a Cj detonation or a weak deflagration when you are looking at these numbers right.