 In the last class we have seen how ISP and non-dimensional thrust depend on efficiency as well as ? B and ? 0 okay in this class we will go further and find out what happens can we get a ramjet which can start below mark number one is that possible or a subs subsonic ramjets are they possible and then we will also look at whether ramjets are self starting we know for sure ramjets are not self starting whether our equations will bring it out is what we are going to see fine and we will also look at what happens to thrust per unit mass flow rate and other things as we go along in this class now if you remember we had derived expression and said that 1 plus ? – 1 by 2 m 4 square must be equal to ? into okay this was the expression that we had derived in the last class now if you put ? is equal to 1 that is ideal case if we look at an ideal case okay what happens here we get this relationship that m 4 must be equal to m 0 does it mean that at the exit mark number is the same as the inlet mark number then it won't produce any thrust is that the case what do you think will happen we are looking at an ideal case and this case exit and inlet mark numbers are the same so do you think it will produce thrust or not produce thrust yeah you are right if you look at mark number mark numbers might be the same what we are interested in us exit velocity and inlet velocity okay so mark number being the same doesn't mean anything because if you know that at the exit of the nozzle temperatures can be 500 to 1000 Kelvin greater than inlet so if you have the flow through the nozzle wherein the temperatures are higher and the speed of sound at that location will be higher and therefore you will be able to still produce thrust if efficiencies are 1 now we derived expression for non-dimensional thrust as okay this was our non-dimensional thrust equation now in this if we put ? is equal to 1 this simplifies to okay so for efficiencies being all 1 unity ideal system you get this expression for non-dimensional thrust and similarly one can get an expression for ISP now if you look at this expression for non-dimensional thrust what you see here is if ? B increases what happens to the non-dimensional thrust ? B remember is under the designers control that is you can choose what is the exit temperature at the exit of the combustor by burning suitable amounts of fuel there is an upper limit that is if you are using if you are using kerosene air there is an upper limit of 2300 Kelvin which is this stoichiometric combustion temperature but other than that you can vary the air fuel ratio and get different temperatures this number is under your control right and one it is very obvious that if you increase ? B you get a larger F by m.a a not now what this value F by m.a not indicates is that you will see as this number is higher the size of the ramjet becomes smaller so larger value of means smaller size of which is desirable because your drag will then be smaller okay so you need a larger value of this and if ? B increases you will find that as a consequence F by m.a a not also increases okay now when we were discussing ramjets earlier in the introductory classes I had made a point that ramjets are not self starting that is they need something to take them to a slightly higher Mach number and then they will start from their own let us see if we can bring that out from this equation if we put m.0 equal to 0 here what happens m.0 is the Mach number at the inlet if we put that to 0 that is if you are holding the vehicle stationary what happens to the thrust this goes to 0 so not self starting or in other words static thrust is 0 so if you put the ramjet on a thrust stand and you do not do you do not give the flow that is you do not induce the flow yourself then it would not be able to produce any thrust even if you switch on the fuel flow rate okay we know now that it does not produce any static thrust is there any value of Mach number at which it will start producing some thrust that is our next exercise we know that static thrust is 0 here let us now find out what happens at what Mach number can we get some meaningful thrust okay now we know that efficiencies if you remember efficiencies will always be less than 1 and when we multiply all of them the product range between product that is ? range between 0.84 and 0.97 right so efficiencies will always be less than 1 so if you incorporate it and try and see what happens at what Mach number will we get some thrust okay to derive that we need to look at this expression here okay this is the expression and we know that if it has to produce positive thrust this number should be greater than 0 right so if it has to be greater than 0 then obviously the product of the first two numbers must be greater than 1 or if you look at the terms in the bracket this should be greater than 1 so we will put that down I will also include the 1 plus f term here the remember we had neglected this term for our convenience okay so for positive thrust 1 plus f should be greater than 1 okay now we can square both sides and cross multiply and we will get this condition that should be greater than ? V okay even though f is smaller here this quantity can be slightly larger therefore I have taken that into account in this equation okay now this is a slightly cumbersome equation to solve and bring out what should be the value of Mach number at which it will give positive thrust so the easier way is you plot a graph and then try and find out where it leads you okay and that is what is done in this plot here what is plotted as non dimensional thrust variation with flight Mach number that is as you increase m0 what happens to f by m.a0 you notice that there are four lines here each corresponding to a different value of efficiency this is for ? is equal to 1 so you have it starting from 0 itself okay whereas all the other lines are clubbed somewhere close to 1 where efficiencies are around 0.86 okay so if you have a non unity efficiency that is for real systems you notice that it is very difficult to have a subsonic ramjet okay whatever be the ? B that you can go up to right ? B is the ratio of combustion exit temperature combustor exit temperature to the inlet temperature even if you add more heat even if you keep on adding more heat it does not matter we bring very ? B from 6 to 10 almost by a factor of 2 even if you add more heat it does not matter you still cannot have something like a subsonic ramjet okay this is purely determined by efficiencies for non unity efficiencies you cannot have subsonic ramjets and what you will notice here is that all these systems have their thrust to or the non dimensional thrust is maximum as somewhere between 2 to 3 okay so it is a maxima for f by a Mach number range of 2 to 3 we need to take this result a little more carefully what we have looked at in this analysis is this is only the thrust produced by the ramjet we have not accounted for the net thrust that is thrust- the drag right there is a drag component also that determines whether the vehicle will move forward or not and that will make it go even further right if you look at this graph in this graph at this Mach number f by m.a0 is this much there will be drag right so that drag will bring down this value and probably you will start producing net thrust somewhere at a slightly higher Mach number right so typically you will not find any subsonic ramjets that is the underlying message from this graph okay when we consider net thrust that is thrust- the drag portion then the starting Mach number will be higher we cannot have subsonic ramjets for efficiencies less than 1 okay now in this graph we see that f by m.a0 varies in this fashion and if I were to look at ISP variation remember ISP by a0 is nothing but m0 into okay now ISP what would be a preferable thing having a larger value of ISP or a smaller value of ISP if you remember SFC we need a smaller value and ISP being 1 over SFC we need a larger value of ISP if you notice here you have ?B appearing here as well as here right in two places in this expression and here it is under root sign and here it is not under root sign okay so what do you expect to happen if ?B increases what happens to ISP ISP should decrease okay and what should happen with efficiencies if efficiencies are less than 1 what should happen to ISPs it should also again decrease and the same thing is plotted here in this graph what you see here is ISP on the y-axis versus Mach number the flight Mach number for an efficiency of 0.86 with different ?Bs as ?B is increased from 6 to 10 you see that the ISPs are lower okay so which means SFCs will be higher or fuel consumption will be more which is not a desirable feature so you would like to operate such that ?B is lower so that your fuel consumption is less but then if you go back to the previous curve you will find that if ?B is less your F by M.AA0 will also decrease or the thrust will also decrease so you should design such that this is a large quantity so that your size becomes smaller okay and then the other thing will automatically get fixed okay now we looked at how the non-dimensional thrust varies with ?B let us look at what happens with M0 we have already seen that in graph let us try and bring out if we can do some analysis and find out what happens with Mach number Mach number if you notice here it appears this is the expression that we are looking for it express it you find Mach number in M0 and ?0 it appears in both places so it is not obvious as ?B is we noted that with the increase in ?B it is this becomes smaller which is what we found in the curves but Mach number is not such a straightforward relationship let us try and find out what happens there that is what we are trying to find out us is there an optimal Mach number at which if we fly it seen that it should lie somewhere between 2 and 3 can we bring that out through some analysis here so what do we do we take a derivative with respect to M0 so firstly we will assume ? to be 1 so that we do not end up with a large or a cumbersome equation so if you look at ? is equal to 1 the expression for F by M0 a0 would be M0 into under root ?B by ?0 minus 1 right so now if we take a derivative of this with respect to M0 let me also include 1 plus F now if we take a derivative we will get 1 into the rest of the things 1 plus F by ?0 minus 1 minus half M0 into 1 plus ?B ?B does not vary with M0 so ?B does not get changed here so ?0 to the power of 3 by 2 into ? minus 1 M0 okay and for a maxima this must be equal to 0 so if we equate this to 0 then both these 2 must be equal we will get 1 plus F this must be equal to 0 under root ?B by ?0 minus half into ? minus 1 M0 square okay what is this quantity here what is the definition of ?0 ?0 is so what you have here is ?0 minus 1 so if you put that you will get 1 plus F ?B minus ?0 into and do the simplification that you will get this ?0 minus 1 divided by this must be equal to 1 or ?B by ?0 into this becomes ?0 minus ?0 plus 1 so ?0 cancels out you get 1 by ?0 is equal to 1 ?0 to the power of 3 by 2 must be equal to okay this is the expression that we get for maximum value of F by M.A0 occurs when ?0 is in this this is the value of ?0 that we have now you plug in some typical values we will find out what we had discovered earlier that it ranges somewhere between 2.52 so if we take TT3 to be 2300 Kelvin which is what you will get as the adiabatic flame temperature if you use kerosene air system at stoichiometric condition and I will take T0 to be 230 Kelvin that is as you go higher in altitude temperature drops so at some particular attitude you will get this temperature and therefore my ?B will be 10 and if I use 7 okay you will find that ?0 you will get as somewhere between 2.25 to 2.15 this depends on what value of F you consider and correspondingly this will also change a little okay and therefore M0 to be somewhere between 2.42 to 0.5 so you will notice that most systems most real systems although all this analysis we have carried out with ? to be 1 even if you put in the efficiencies we have seen in the graphs that Mach number range for high value of F by M.AA not ranges between 2 to 3 which is where you will find most operating systems most systems would want to go to that condition and then fly in that Mach number range so that you have a larger value of F by M.AA not okay this finishes our discussions on ramjets now let us move on to the next engine that is the turbojet okay turbojet has a slightly involved system it has a compressor and a turbine so we need to account for that here okay now when we look at turbojets there are two things that we can look at one is when the nozzle is choked and the other is when the nozzle when the flow through the nozzle is optimally expanded firstly we will consider the case wherein it is the flow through the nozzle is optimally expanded and we will try to find out what derive what equations we will get now turbojet similar to a ramjet will have an intake and then we will have a compressor which is connected to a turbine to have a combustion chamber in between and then you have a jet pipe or an afterburner and then a convergent nozzle okay so let me again call these stations as 1 2 3 4 5 6 and 7 this is the 0 to 2 is the intake 2 to 3 is compressor 3 to 4 is combustor this is the turbine this is the afterburner and then the convergent nozzle as I said earlier we do not use typically CD nozzle the only aircraft or a real system that has a turbojet which uses a convergent divergent nozzle is the what is the engine Olympus engine right on what is that air on quad aircraft okay so other than that typically only a convergent nozzle is used because the pressure at the exit of the turbine is very low and it would not be useful if you have a convergent divergent nozzle and therefore only a convergent nozzle is used now here we will take up the case where P7 is equal to P0 okay that is optimally expanded flow notice here that you have compressor extra and the turbine extra so we need to define certain other parameters that is we will define compressor pressure ratio I will call it by C I see is nothing but Pt3 by Pt2 and T again here indicates stagnation conditions from this we can define Tc that is nothing but Tt3 by Tt2 which is I see to the power of ? – 1 by ? okay and similarly I can define a turbine pressure ratio so from this I can define Tt which is nothing but Tt5 by Tt4 so now with this and we know that from our previous calculation ?0 is nothing but Tt0 by T0 that is 1 plus ? – 1 M0 square fine and based on this I can get M0 is equal to under root 2 by ? – 1 ? 0 – 1 the other parameter that we need to define is ? B in this case I will define ? B is Tt4 by T0 okay so ? B is nothing but Tt4 by T0 which I can rewrite as Tt4 by Tt3 into Tt3 by Tt2 into Tt2 by Tt0 Tt0 by T0 what is Tt4 by Tt3 this I will call it as Tb similar to what we had done earlier I will call this as Tb and Tt3 by Tt2 again we have it here it is ?c and Tc sorry and what is Tt0 Tt2 by Tt0 if we assume isentropic processes this will be 1 this is flow through this is flow through intake and just like in the previous case if the process is isentropic we can get this to be 1 and lastly this is nothing but ? 0 okay so I get my expression for ? B or Tb I can write it as now we want to derive expressions for thrust so the thrust equation is f by f is equal to m dot e into in this case the exit velocity is at v7 okay and intake is at 0 so you get v7 into okay we had assumed p7 is equal to p0 and therefore this goes to 0 and we will also make the other assumption that f is very much less than 1 this is more true in a RAM in a turbojet compared to a ram jet because in the main combustor it will be much lower than 0.067 because you are worried about the turbine inlet temperature so you wanted to be much lower than these stoichiometric condition so f is less than 1 so I can neglect this part and I get f is equal to m dot a v7 minus v0 and we will do the similar procedure as we did last time so I can write this as m dot a a0 m0 into m7 by m0 into under root T7 by T0-1 again we have used the relationship for speed of sound here okay and we have again made the assumption that ? and R are the same for exhaust gases as well as incoming air so assuming ? and R to be same for gases and air okay so we get this relationship now again we need to look at these two ratios okay or we can write the non-dimensional thrust as we can write the expression for non-dimensional thrust as m0 into m7 by m0 so we are again left with the task of trying to find these two ratios again we will follow a similar procedure of cascading temperatures and cascading pressures right now what I want you to think about is all the while I have been telling you that you are looking at flow through intake and I said we are looking at isentropic conditions the previous case that we took up ramjet we looked at all the temperature ratios and efficiency is never figured in that place right why does efficiency only figure when we are looking at pressures and why does not it figure here now in the previous case also we looked at a realistic system right so we cannot look at realistic systems and say the flow through the intake is isentropic so what really is the crux there why are we taking this is one and why do we take the other thing wherein when we look at pressures we are looking at efficiencies here okay we will discuss this in the next class okay thank you very much.