 In the last class we had seen how efficiencies affect the performance of gas turbine engines that is turbojet engine in all these classes in all our analysis till now there is an assumption that we made that CP is constant ? is the same right and if we take f into account also fuel a ratio we have assumed to be very much less than 1 and therefore we have neglected it and we have done all this and our analysis does not seem to include if there is bleeding from some of the compressor stage after the low pressure low pressure compressor stage if there is bleeding how do we account for it okay if CPs are different if ? the ratio of specific heats is different how do we account for that now what we will do is we will do a step by step procedure that is we will go from the inlet to the outlet and we will put down what do we know and what do we need to get and try and see if we can get it even with all these complexities that are there in an actual engine the actual engine it is not as simplistic as we had assumed that efficiencies are all 1 and all that so we will see how we can still analyze an actual engine with this firstly the TS diagram would be this is for a cycle with all efficiencies being 1 now if you have an actual cycle this will be 2- 3- the dotted one shows the actual cycle right now what we are going to do in this analysis is if we know conditions at the inlet can we get whatever parameters we were looking for that is if we know P0 T0 M0 this is the inlet conditions and then let us say we know the compressor pressure ratio ? C then the other parameter that we know is the turbine inlet temperature okay then the exhaust area a7 and all the efficiencies efficiencies I will call it as ? I if we know compressor efficiency intake efficiency then efficiency of the burner nozzle turbine efficiencies all this if I know it I will put this in ? I okay now if we know this can we calculate now from this what is the parameter that we are of interested in F that is the thrust of the engine then the ISP of the engine then the other non-dimensional quantities that we looked at so we will try and see how to get to this knowing this one okay so as a first step we also need to know in addition to this we need to know what else we need to know thermodynamic properties right like thermodynamic data that we require our CP of turbine then CP of compressor then CP of burner then heat of combustion of the fuel that is Q right then we need to know ? right if we know this and if we know this can we calculate thrust ISP non-dimensional thrust and non-dimensional ISP this is under the condition wherein you could have bleed you could have a patience is being different from one and also all these CP is being not equal to 1 and ? being also different so as a first step step 1 I know T0 I know M0 what can I calculate from these two stagnation temperature right I can calculate T0 and from that I can calculate ? 0 okay so I can write ? 0 if you remember was 1 plus ? – 1 by 2 M0 square I know M0 so I can get ? 0 okay right and I can also from this get T0 which is nothing but T0 x ? 0 and similarly I know P0 so and I also know ? 0 so I can get P T0 the stagnation pressure at inlet as P0 x ? 0 to the power of ? by ? – 1 okay. So we were able to calculate the stagnation conditions here fine now in the next step what happens between 0 and 2 this is the intake here stagnation temperature remains the same stagnation pressure changes because of diffuser efficiencies or efficiency of the intake right. So if I know all the efficiencies fine that is diffuser efficiency burner efficiency compressor efficiency turbine efficiency and then nozzle efficiency I can calculate all these parameters so let us do that I need Pt2 that is pressure at this point so I will get that by ? diffuser x Pt0 okay and I can get Tc is equal to ? c to the power of ? – 1 by ? so I know Pt2 I also know Tt2 is nothing but Tt0 so I know Tt0 so I know fresh stagnation quantities at 2 so I can calculate quantities at 3 fine I know conditions at 2 now I can calculate at 3 by Pt3 is nothing but ?c x Pt2 fine I know Pt2 I can calculate Pt3 and I want this temperature right for an actual cycle so I can calculate Pt3 dash as equal to okay so I know Tt2 I know Tau C I know ?c so I can get this TT3 dash so I know now the conditions at the inlet of the combustor so next I need to know Pt4 what is Pt4 there is a pressure drop as you go from the inlet of the combustor to the exit because of a rally process wherein you are adding heat therefore pressure will drop okay so I know the efficiency of the burner ?b x Pt3 will give me Pt4 and what is TT4 TT4 is nothing but the turbine inlet temperature that is a given condition so now we know conditions at this point that is temperatures and pressures we know efficiency of the turbine okay and we also know the compressor pressure ratio so using this we can calculate what is the condition at 5 okay what we have used here is that the CPs need not be the same okay CP of compressor CP of turbine that is the burnt gases need not be the same so we have accounted for it and we have also taken a condition wherein F need not be very small or if we do not want to neglect that part or be happy with that error we can take into account here knowing TT5 dash I can calculate TT5 as TT4- efficiency of the turbine okay and from this I can calculate the PT5 PT5 is nothing but this quantity is Pt sorry this quantity is Tt this whole thing is Pt okay so Pt4 x Pt is what gives me Pt5 now having known this Pt6 in case we are using the after burner okay even otherwise also there is an efficiency there is a pressure drop across the jet pipe so we can take that into account here into Pt5 gives me Pt6 and TT6 is equal to TT5 right if there is no after burner if we switch on the after burner then we need to take that into account here that will be specified in addition to TIT that will be specified to us that will be known to us so if it is given then we have to take into account that part also and in the next step we calculate what happens through the nozzle okay so what we have done is starting from here and knowing efficiencies and CP we have been able to calculate all these parameters fine now we need to calculate thrust right there are two conditions that can exist one is depending on the nozzle pressure ratio we could either use a convergent divergent nozzle wherein we will assume P7 is equal to P0 or even in a convergent nozzle a special case P7 is equal to P0 or nozzle is choked let us first take the case where P7 is equal to P0 okay this can be for both CD as well as convergent nozzles. So if P7 is equal to P0 then I can calculate Mach number M7 is equal to so I know Mach number at the exit what can I find out I need also a7 right if I know a7 then I can calculate velocities right so a7 is nothing but under root ? RT7 but I do not know T7 as yet so let us get T7 as what do we know TT7 by T7 is equal to or T7 is equal to okay so I know M7 I can get I know TT7 so I can get T7 and since I know T7 I can substitute it here I get a7 now knowing Mach number and a7 I can calculate the velocity at the exit that is V7 I also need to calculate the mass flow rate okay. So mass flow rate we can mass flow rate leaving the I can do two cases one is m.a is equal to m.a x 1-f is equal to ?7 if I do not want to neglect F I can write this expression I know this to be P7 by RT7 x V7 a7 now here I know all the quantities I can get m.a x 1-f right I still do not know what is F we will get to that shortly how do we get F if you remember when we did ISP calculations what did we do energy balance across the burner so we will do that right this is the expression that we get if we do energy balance across the combustor we know this quantity m.a x 1-f what we have just now calculated this will be given to us these two things are what you have seen earlier TT3 dash is what we calculated TT4 is given to us so we can calculate m.f right now F is nothing but m.f by m.a so I know m.f from this I know m.a plus 1 by 1 plus f and using these two I can calculate m.a right that clear so from here we get m.a now I know m.a m.a x 1-f all these quantities so my thrust equation is F is equal to m.a x 1-f V7-V0-A7 P7-P0 this goes to 0 because we have assumed P7 is equal to P0 so here we know this we know this we know this now what we do not know yet is V0 now I know m.0 right and I also know T0 so I can calculate V0 firstly V0 is nothing but E0 M0 that is equal to okay so we know about T0 and M0 so I can calculate V0 so I will know everything here right in this equation all these quantities are now known and therefore I can get thrust and once I get thrust I can also get ISP very easily ISP is nothing but thrust per unit mass flow rate of fuel I know mass flow rate of fuel I have calculated here I know thrust so I can get ISP now what are the other quantities the non-dimensional quantities I know A0 also so I can get ISP by A0 I also can get m.f by m.a A0 so all these quantities we can get by doing this analysis going step by step one can also look at there are cases there are certain engines wherein a portion of the compressor compressed air is bled to cool either the turbine blades or you use it for air conditioning the aircraft certain quantity of the compressed air is taken out especially after the low pressure stage so you can then suitably write the equation for power balance across the turbine and compressor taking that into account because there will be a reduced mass flow rate through the second stage of the compressor and also through the turbine so going by the step by step procedure you can calculate that and account for it also right so in this way we can do even for a realistic engine all these calculations not as if that what we did was something that is not applicable to any realistic engine this is applicable to any kind of realistic engine with bleed and other things also okay now we have looked at ramjet we have looked at turbofan sorry turbojet and turbojet we have considered all the cases that is after burner on then water methanol injection converging sorry one minute I need to still we have only considered one case P7 is equal to P0 we need to consider the other case wherein we have choke condition so let us do that first case when nozzle is choked we know that M7 is equal to 1 so if M7 is equal to 1 then I know that T7 is equal to 2 x T7 by ? plus 1 right so I also can similarly get P7 as now I know T7 using this I can get a7 is equal to and I know M7 is equal to 1 so I can get V7 is equal to M7 x a7 and similarly m.a x 1 plus f is equal to row 7 by RT7 a7 V7 and a similar expression for m.f q is equal to m.a x 1 plus f see combustor I know from here I can get f which is nothing but m.f by m.a I know m.a I know m.f I know f I know V7 what else do we need V0 we know is nothing but M0 a0 that is M0 x so knowing all this I can get again f is equal to okay I know P7 I am given P0 m.a is known V7 is known V0 is known a7 P7 and P0 so I can get thrust from here I can get ISP as f by m.f and the other non-dimensional quantities that we wanted so we have looked at the cases wherein for ramjet as well as turbojet we have looked at cases where efficiencies are equal to 1 not equal to 1 and for a turbojet we have looked at case wherein with after burner without after burner choke nozzle optimally expanded flow then we have also looked at case with water methanol injection all this we have looked at now let us look at the next two classes of engines that is turbofan and turboprop turboprop is very simple we do not need to do any analysis except that all the power that is developed in the turbine must be equal to the power of the compressor as well as the power of the fan propeller right so if you do that you will get the turboprop engine that is not a big problem but a turbofan engine needs some effort so let us look at the turbofan engine first so this is a turbojet engine and to this we add the turbofan part both of them all of them are mounted on the same shaft right now are typical conditions where this was 0 after the intake we had to write this was 3 4 5 6 and this is 7 whatever we had done for the turbojet remains as this now for the turbofan you have the fan in front also and there is a certain amount of air that is bypassing and is going through only the fan portion okay so if you look at condition after the fan we will call it as 8 okay and we will call the exit condition here as 9 at the exit of the fan as 9 the bypass exit as 9 okay so you have 0 to 2 2 to 8 8 to 9 that is the additional part in the turbofan engine the rest of it is a turbojet engine right so how does the thrust expression look like for a turbofan engine f is equal to m.a x 1-f V7 V8 V9 V7- V0 plus a7 P7- P0 all this is turbojet part okay now in addition to this you will also have alpha x m.a V9- V0 plus a9 P9- P0 this is because of the fan or the bypassed act alpha is the now we will make an assumption that the flow is optimally expanded through both the nozzles there are two nozzles now one is this nozzle and the other one is this nozzle so we will make an assumption where we say flow is optimally expanded so therefore we get P7 is equal to P9 is equal to P0 okay so this part remains the same as what we did for turbojet analysis we now have to only look at this part because of our assumption here this goes to 0 this goes to 0 so I have to only take care of this term all the rest of the analysis that we have done for this for a turbojet with optimally expanded flow is valid okay so in order to do this what do we do what are the things that we need V9 by V0 is what we will get so we need T9 by T0 and P9 by P0 that is from here we will get M9 by M0 so knowing these two we can also get this portion non-dimensionalize this portion and get this value okay so what is T9 by T0 you have T9 by TT9 TT9 by TT8 okay fine now I know this quantity is ?0 fine this is ratio of static to stagnation conditions so I will put this as 1 plus 1 by ?-1 M9 square next is flow through this duct here okay this is similar to the jet pipe so if we assume all efficiencies to be 1 which is what we will do so here we are assuming also ? is equal to 1 okay so TT9 by TT8 would be 1 and TT8 by TT2 is process across this fine I will define a new ratio that is I will define ?f is equal to TT8 by TT2 and ?f is equal to PT8 by PT2 so this would be ?f would be equal to ?f to the power of ?-1 by ? okay fine so coming back here I will get this is ?f into this is TT2 by T0 it is 1 so I get ?f ?0 by 1 plus ?-1 by 2 M9 square okay when we cascade pressures we will get this ratio also so let us do that this will be P9 by PT9 into PT9 by PT8 by PT2 by PT0 by P0 okay this quantity is nothing but ?0 to the power of ? by ?-1 this is equal to 1 and what is this this is ratio of static to stagnation what about this PT9 by PT8 this is 1 right so this is 1 and PT8 by PT2 this is ?f and again PT2 by PT0 is 1 because we have assumed all efficiencies to be 1 into ?0 to the power of ? by ?-1 so if I change this to Tf I will get all of them in powers of ? by ?-1 so I will get 1 plus ?-1 by 2 M9 square is equal to Tf ?0 okay so I can come back here and I know that this is again ?f ?0 divided by ?f ?0 so this is 1 which means that if all efficiencies are 1 T9 will be T8 right because here in the intake you are compressing it then you have a fan to increase the pressure and again you are expanding the flow if there are no efficiencies involved there is no heat addition here so T9 must be equal to T0 okay and I also know that 1 plus ?-1 by 2 M0 square is equal to ?0 so using these two I can get the ratio of Mach numbers which will be M9 by ?f ?0-1 divided by ?0-1 so finally I can write the expression for thrust in non-dimensional thrust you do not need this f is less than 1 very much less than 1 so this part is the turbojet part that we had already seen Tau C Tau T ?0-1 okay this is the expression that we get now we need to do the compressor turbine power balance because Tau f Tau C and Tau T are all related okay so let us do that turbine compressor and fan plus a M.A TT 8- TT 2 if we deduce from this similar to what we had done if you remember we had found an expression for Tau T in terms of Tau C by canceling out M.A with assuming that f is very much less than 1 and CP being the same we can reduce this to Tau T is equal to 1- ?0 by ? B Tau C-1 this was if only there was compressor turbine power balance in addition you have the other term – ? okay so if we plug back this into this expression here we will get the overall expression for thrust non-dimensional thrust okay we will stop here and continue in the next one thank you.