 Good morning in the last class we had derived the expression for the unsteady variation of pressure in the rocket motor right we had considered the mass balance in the rocket motor to derive this equation in this class let us look at how we can use that to predict what is the time required for ignition what is the time required for extinction okay till now we had looked at only the steady part that is the burning in the steady portion right this class let us look at the other two unsteady portions that is namely ignition and extinction if you look at a typical thrust time curve or a pressure time curve this is ignition time and this is extinction time now in this class let us try and use the relations that we had obtained in the last class to find out how these vary okay if you remember we had derived the equation for pressure variation as if let us say the motor is operating at the equilibrium pressure itself let us say it is operating somewhere here then this PC this PC is a time variant term okay there is not an equilibrium quantity this PC will also be equal to the PC and as a consequence these get cancelled out and you will get this DPC by DTS 0 okay so in this equation we had identified the VC by AT as L star and this entire term as 1 by PC right 1 by characteristic time so we can write this as now let us say how to use this to get what is the time required for ignition okay and what is the time for extinction before we go there we should understand this that why is this important in a rocket motor okay ignition delay let us say if it is a very long time then also it is a problem it is a very short time then you can have a very high pressure burst in the beginning and then therefore affect the structure okay you could actually burst the motor in that time so either either of the two is not what we want we want to operate it in a very narrow window okay typically even in a large motor the ignition should be over within 100 milliseconds okay 100 milliseconds is it time that we are looking for so let us try and derive the expression for the ignition time now during ignition if you look at this portion this is a very small time as I said it is less than 100 milliseconds right so if you are looking at that kind of a time scale one can make this assumption that the chamber volume and the burning surface area and C star do not change in this period okay VC is chamber volume VC changes only because of burning in 100 milliseconds it does not burn too much so VC and AB will be nearly the same and C star if we assume we will assume constant I can write this term VC equilibrium 1-n as follows KN is equal to AB by 80 I should also add here the throat area also does not change with time in this case right during the initial phase of ignition it is a very small time and we are not looking at hundreds of seconds so the burning the throat area also does not change so in this case KN will be also constant this entire term is a constant and I can absorb it as alpha I will call alpha is equal to rho p KN AC star so if I want to find the time for ignition all that I need to do is integrate this equation now the limits of integration need to be identified okay typically the igniter is supposed to give something like 30% of the pressure rise let us say this is this equilibrium pressure is 100 bar the igniter should give something like 30 bar on X1 okay so around 30% of equilibrium pressure so if we integrate this we will get T ignition is equal to integral okay this is the 1 by TC into TC into this is the equation that we need to solve and alpha we have identified what it is so now we need to play a little bit of jugglery here if you look at the denominator we can express this as let us say X is equal to PC to the power of 1-n okay then what will be DX DX will be 1-n into PC to the power of – N DPC this equation I can rewrite this as T ignition is equal to PC PC to the power of – N DPC then I will be left with alpha – PC to the power of 1-n okay so if I use this here this is nothing but the same equation rewritten if I now use this substitute this here I will get I can change it to the X variable integral X ignition to X equilibrium okay this is nothing but DX by 1-n so I will get TC by 1-n and this is alpha – X okay now we can see that this is in the you can integrate it and get a logarithmic function so I will get alpha – this is the form we will get please note alpha is nothing but this one or this is also equal to right alpha is nothing but this this is also equal to PC equilibrium to the power of 1-n now if you put that back in this equation here you will find that the denominator is going to 0 or the time will go to infinity okay so you cannot have infinite ignition time so therefore what is usually done is we try to reach something like 95% of the equilibrium pressure what is the time so we calculate time required to and that will give us a realistic ignition time okay so this is how we calculate ignition time now if we want to look at what are the kinds of ignitors that are used in the industry yes I said 30% of the pressure rise will have to be given by the igniter itself that pressurization is done by the igniter itself if you have the chamber volume 30% of that initial pressure rise has to be given by the igniter itself when the igniter burns it should fill that volume up to the pressure of typically something like 30% so it usually starts from somewhere here so we are looking at this rise only okay now let us look at what are the kinds of ignitors that are available okay there are two kinds of igniter one is called as pyrotechnic pyrogen typically the pyrotechnic igniter is based on gunpowder it is a derivative of the gunpowder and this is nothing but a composite propellant 30 to 40% metal loading if you remember when we discussed about composite propellants we said it has typically 18% of metal loading here the metal loading is increased to something like 30 to 40% so that the way this operates is it is not only the hot gases that come out but also the hot hot metal particulate matter they go and embed themselves on the solid propellant and they start a local ignition process okay so that is why you have a larger fraction of metal content in these propellants and usually they have very high burn rates if you look at very small motors then pyrotechnic is used medium sized you have a pyrogen igniter and in large solid rocket motors like PSLV stage 1 etc what you use is you have this pyrogen igniter is moving on rails from the head end to the nozzle end so as to ignite the entire motor okay now if you look at what happens when the igniter is switched on the if you look at a rocket motor the igniter is placed at the head end okay and as I said it also has particles that go and embed themselves on the propellant when these particles embed themselves on the propellant they cause local ignition in addition to that there is a hot gas flow in the chamber okay so both of these lead to ignition and if you look at the igniter igniter will have something known as both the kinds of igniter will have something called as quib which is made of this is sensitive to current and is usually made from lead or mercury filminate so when the current is supplied this is sensitive to that this starts to burn and then ignites the rest of the igniter and then this igniter will then ignite the rest of the motor okay as I said if you look at a motor like this we said that the ignition time needs to be even for a large motor less than 100 milliseconds right and I said in the beginning that if we let us say do not do this in the 100 milliseconds then what happens if there is a slower ignition then we could get into combustion instability could set in sorry one minute I said this has to be completed within 100 milliseconds now let us say we have a certain igniter mass right if we put in the required igniter mass we can achieve this within a narrow band let us say we put something more than what we should have put okay then what happens there are two things firstly if you look at a motor of this kind okay a port motor if you have a very large L by D right if you have this length of the motor to I will consider this diameter as important and not the overall diameter primarily because this will be the port it gases that are generated here if you remember this equation that we had derived DPC by DT is equal to 1 by TC a this minus portion what is it due to this is because of flow going out through the nozzle this portion is because of burning right mass addition due to burning and this is because of now what happens it takes time for this to occur right if the length is especially large length to diameter is very large then it takes a longer and longer time for the flow to go out and therefore if you look at this equation the term that is contributing in a negative sense is not going to set in for a long time as a consequence you will have in terms of pressure versus time let us not worry about this portion this is going to increase in this direction with increase in L by D right because it is going to take a longer time for the gases to go out and therefore the pressure keeps on rising and this pressure overshoot will become higher and higher and at some point could be detrimental to the motor in some cases of large L by D motors this is known to be one of the modes of failure of motors so one needs to take care of this so as to provide ignition such that this does not happen also the other trouble is with regards to ignition if you look at this figure this is of a simple port yes I am only looking at this it depends on the volume to but if you have the same volume and if you have a long pipe right if you have a stubbier volume the same one and a long pipe then if you look at both of them this is what you will find right that is what I pointed it out this is also in some sense similar to if you have a tank over a tank and if you have a very long pipeline the changes at the other end will be felt after a long time right so that is the reverse problem that is happening here right coming back I was talking about this motor here if you look at the trouble with ignition this is a very simple motor right we have a cylindrical grain but usually we talked about this earlier if you want to get the required thrust time curves in the motor then the grain has to be either a star grain or a phenoscel or something else right which means that you are going to have cuts here and cuts here and you have to ensure that the hot gases or particulate matter flows into all these crevices and ignites the entire motor in the very short time that is available which is not very easy actually okay. So and in addition let us say due to some reason we do not supply the pressure that is required by the igniter let us say we fall short then it is known to go into an instability mode if you do not supply it the required pressure then it goes into the instability mode so either if you give it more pressure you could have a hard start like what we talked about or if you give it a lower ignition it could go into instability mode so we need to have the right amount of ignition energy so as to provide typically something like 30% of the initial pressurized as well as provide a smooth ignition now we can calculate what is the amount of igniter required based on this formula the igniter mass is given by 1-PC ignition now if you look at this relationship the SF stands for solid fraction it is assumed here that the solid fraction does not give rise to the pressure right so you have to count that out and then you will get the pressure rise that we are looking for this has to be typically 30% of the equilibrium pressure this is the chamber volume and this is ignition the temperature at the end of ignition okay so using this we can calculate what is the amount of igniter that needs to be carried so as to get a good ignition now let us look at the other end of the spectrum that is the extinction part okay we have looked at the ignition part the other end of the unsteady process is the extinction let us look at that now if you remember when we discussed about burning of the web and other things we discussed something known as Liver loss and I said this is something called unburned propellant okay now let us find out why certain portions of the propellant do not burn and stay like that although it was burning up to some time and suddenly why does it quench itself okay now if you look at a missile or a launch vehicle both of these have the objective of placing its payload at a particular orbit with a particular velocity right so as I said earlier that it is more so with missiles than with launch vehicles the temperature varies from place to place as well as day and night okay and as a consequence the burn rate also varies now with all this you still have to place it into the orbit at a given velocity and at a given altitude which means that if you could have in some sense a possibility of cutting off the thrust okay then it would be a lot more easier it is very easy in liquid rockets that is you just stop the supply of the liquids or also in hybrids one of them is a liquid so you stop the supply of the liquid you will be able to stop the combustion process but how does that happen in a solid is what comes under extinction but before we do that let us find out what is the time it takes for the extinction after extinction the depressurization right if you look at the pressure time curve we are talking about this time okay what is observed here is that the combustion is completed we will see why should the combustion get completed at that point a little later now if the combustion is completed what happens to the burning surface area burning surface area we can assume it as 0 a b is 0 then what happens to a in the equation that we had derived in this equation what happens to a a is nothing but to p a b by a t a c star if a b goes to 0 a also goes to 0 so a equal to 0 then if you look at this equation what do we have this term goes off and we are left with minus pc by tc so we get the equation as now we can integrate the equation and find out what is the time it is a very simple you get ln pc must be equal to minus t by tc please remember that the tc there is a constant okay so you will get this which means that pc follows an exponential curve pc we can write it as pc equilibrium exponential of minus t by tc that is the pressure here decoys in an exponential fashion and please remember this equation is valid up to what point of this drop up to the point where the throat is choked okay if you remember the equation that we had is for a choke nozzle right so this is valid up to the condition where the throat is choked now this time is of importance because if you look at either a multi-stage launch vehicle or a multi-stage missile when you cut off the stage right if it is still burning like this because the weight is very small it could accelerate and come back and hit the mother vehicle the remaining portion of the vehicle so you would not want that to happen and you would want to know what is this time in this time it can still be thrusting right the pressure is decaying so it could still be thrusting and you do not want it to come back and hit the vehicle right so in that sense this time is of importance now coming back to the question why should propellants quench when they are depressurized or why does it quench or why should there be a sliver loss right if you remember our discussions earlier we had said that there will be some portions of the propellant that will remain unburnt if you look at the if let us say this is the cross sectional view of the rocket motor there will be some portions that will remain unburnt now to ask ourselves this question why should it not burn up to this point it was burning right and why does not it certainly or why should it suddenly stop burning the answer lies in this fact that suddenly the burning surface area decreases if you look at it it was earlier occupying the entire this entire circumference and suddenly it ceases and becomes only in some portions that you have a propellant right so what happens the burning surface area AB suddenly changes if AB changes suddenly and if you have a propellant with high N or a medium N also I mean burn rate pressure index N is moderately high right not we are not looking at 0.8 or something we are even if it is 0.3 or 0.4 what happens to the equilibrium chamber pressure when this happens when the AB suddenly changes or it is a larger burning surface area and suddenly it decreases to a smaller burning surface area if you look at our expression for equilibrium pressure right so if AB changes suddenly then the pressure also changes suddenly and if AB drops pressure drops suddenly in a sense you are depressurizing the motor okay so people have done this experiments and try to find out what happens in this case what happens when the motor is depressurized or when a propellant is depressurized this is one of the ways in which you can stop the propellant from burning further if you depressurize it prevents the burn propellant from burning or it inhibits the propellant from burning later on let us see why this should happen it is something like this let us say a propellant is this is the burn rate versus time that I have let us say a propellant is burning at some rate here suppose you depressurize it okay then at the lower pressure which you bring it to let us say it has a burn rate something like this right it has been known that if you do this process in a very small time that is if your DPC by DT is large right then the propellant tends to quench if you do it very slowly it will go to this burn rate if you do it very rapidly at some point it could do this okay if you do it further rapidly it is seen to quench like this okay. So there is a critical DPC by DT beyond which the propellant ceases to burn and people have kind of done the study wherein to find out if let us say you have an initial pressure and this is the DPC by DT the curve follows something like this if for an initial pressure somewhere here if you are using let us say this depressurization rate then it will it will quench but let us say if you are for the or it will not quench and let us say if you are this is not quench and let us say you are here this will lead to quenching in a sense this line defines the boundary at which the quenching or the not quenching happens okay so this line defines the boundary at which quenching and no quenching takes place so if you are above that it will quench if you are below that it will not quench and also please remember this quenches still even when there is a burning possible at that lower pressure it is not as if we are taking it below a pressure where the propellant ceases to burn right. The answer for this is in some sense in the characteristic time of characteristic conduction time scale that is the for the solid the characteristic time scale is much much greater than the characteristic time scale for the gas we will discuss this a little later as to why it should be for the solid it should be higher than the gas so in a sense the gas responds very quickly to the changes whereas the solid is very very slow and responding to changes typically it is of the order of a 100 times or 1000 times 100 to 1000 now what happens in this case is if you have these gas respond very quickly let us say we have the situation wherein the propellant is burning in this fashion let us say there is a flame here now if you depressurize this the flame tends to move up because the pressures have reduced and the flame tends to move away from the surface when this happens but the solid tends to feel that it is operating at the previous condition itself and tends to pump in more gases as a consequence you are filling in more inerts in a sense and this causes the flame to move even further away whereas if you look at what happens inside the solid the solid will need more and more heat to operate at a lower and lower temperature so in a sense you are moving away from the situation where at the point where it was burning the solid was getting the ample heat that is required for it to continue burning you are now moving into a situation wherein it needs more heat than what it is getting and the heat is also decreasing so in a sense you are pulling away from the stable point and therefore it tends to quench okay that is the reason in some sense for this quenching we will stop here and continue in the next class the next class we will discuss erosive burn okay thank you.