 In the last class we had seen the performance of a turbojet we derived equations for the non-dimensional thrust as well as we had derived expressions for ISP of a turbojet. Now let us put in some typical numbers and see what is it that we get out of it and we will see some interesting facets of it. So firstly we will look at optimal expansion of flow in the nozzle here there are two conditions that we need to look at one is C level thrust or C level ISP and what happens at cruise conditions so at C level P0 is equal to 1 atmosphere or 1.1 mega Pascal's and T0 would be if we assume an IS ISA atmosphere model then it will be 288 Kelvin and M0 is 0 at C level before take at takeoff conditions and now if we assume the compression ratio pi C to be very low that is pi C to be of the order of around 4 then my tau C would be something like 1.484 and if we take TT4 to be 1000 Kelvin then this makes ? B is equal to something like 3.4 and we know the Q of the fuel this is kerosene so 42 mega joules per kg and let us take the CP of air and flow gases to be the same and let us take it to be 1 kilo joule per kg Kelvin okay. So with this conditions now you have to find out what is it at the cruise altitude of around 11 km so altitude whatever things that are going to change the things that are going to change are these two only atmospheric conditions and at cruise conditions this Mach number will also change so we need to change the only three parameters P0 would be lower and it would be somewhere around 22.6 kilo Pascal's and P0 would be this is we have taken ISC international standard atmosphere so therefore we get this now if we use this and calculate f by m.aa and ISP okay we have all the parameters here to calculate both of them or we will get this kind of a table m. is equal to 0 then I will also put V7 V0 both these in meters per second and lastly ISP in kilo Newton second per kg okay so if we take the C level performance that is at C level m0 is equal to 0 and using this we will get f by m.aa not as 1.8 and this would be something like 612 meters per second and this would be 0 and ISP would be something like this now at 11 km m0 would be 0.8 and this would reduce slightly to 1.76 okay now there are two things to notice one is that irrespective of the altitude you see that f by m.aa not as nearly constant right this happens because of two things one notice that at C level this is large the velocity differential is large okay the mass flow rate is very small velocity differential is large mass flow rate is very small so you end up getting a high thrust because of the velocity differential at a higher altitude the velocity differential is reduced but you have air coming in now at this velocity therefore what you get is a larger mass flow rate so you get a larger mass flow rate with a smaller velocity differential therefore you get nearly the same thrust or same non-dimensional thrust what happens to ISP ISP is come down which means that SFC is gone up if I were to write SFC also here SFC would be in kg per kg or this would be something like 0.8 this would be like 1.06 so we are cruising at an altitude wherein the SFC is low why do we do that I think it would be more beneficial if we were to stay on ground or close to ground and carry out our operation is not from the looks of it what we see is that the ISP has come down or SFC has increased so why do you think if this is happening or this is desirable non desirable what is the opinion you have so one say something so what happens no not really well actually the reason why we operate it at a higher altitude is primarily the drag is less at a higher altitude right you are not looking at drag of the engine per say you are looking at the drag of the entire aircraft wherein when I was discussing about turbofan engines I said G goes in for a bypass ratio of 9 and says that the engine of the nasal drag is very small compared to the overall drag and therefore they go in for a large bypass ratio right now what we are looking at is the drag of the entire aircraft goes down as altitude increases because roading road decreases as you go up in altitude right now density decreases drag decreases but your SFC will increase because what we have set in here is if you notice I did not change this part TT4 TT4 is the same at both the altitudes and if you look at ? B in this case ? B would be because T0 is lower so ? B for TT4 of equal to 1000 Kelvin would be 4.63 that is what we are saying is you need more fuel to bring the air from a lower temperature after compression to 1000 Kelvin because we have set that temperature so therefore we will find that the SFC will be higher right okay and that is the reason we see this now we can look at similar numbers for choke flow through the nozzle the only thing that changes would be I will just put them here itself now here we will assume a slightly larger compression ratio of round 12 which means that Tc would also go up it would be much well can also increase the turbine inlet temperature this means ? B will go up to the rest of the quantities remain the same here again sorry I had written 100 Kelvin here you are not pointed out it was to be 1000 Kelvin earlier and that makes it 4.63 okay and that is why we saw that change in ISP or SFC now we change it to 1200 Kelvin ? B would be 5.55 and this would mean correspondingly all these quantities will change because now we have choke flow in the nozzle the thrust consists of two parts one is the convective flux part and the pressure thrust part so this will again have two parts one point and at an altitude is what we see notice that as you go up an altitude the fraction of the pressure thrust is increasing okay it is a lower fraction here right or if you look at this number it is a lower fraction here compared to a higher fraction here right the overall thrust why does this happen why does pressure thrust part increase as you go up an altitude higher ambient pressure if you see is one-fifth so the pressure thrust part is going to increase if you go up an altitude okay and again you see the same feature here that SFC has increased and that is primarily because you need to heat the air from a lower temperature to a set temperature of 1200 here instead of 1000 earlier so therefore you will find that SFC is higher in this case okay fine okay then we have dealt with two things we looked at optimal optimally expanded flow through the nozzle and we looked at choke nozzle okay now let us look at what happens if you have the after burner on and the other method of increasing the thrust that is water methanol injection let us look at these two methods and how to derive expressions for them okay so firstly let us look at what happens if we have the after burner on I will retain these set of numbers because I want to use this to look at what happens without the after burner with the after burner conditions I will retain this numbers and we will get back to a similar table at the end of the discussion on afterburners now if we look at the TS diagram with the after burner on we will again assume ? to be one okay so you have isentropic compression through the intake as well as through the compressor then you have heat addition in the main combustor expansion through the turbine so this is 0 to so this is the TS diagram for efficiencies one and with the after burner switched on okay now what do we have to do if we have to analyze this okay firstly our expression for f by m.a0 the non-dimensional thrust would be m0 and 2 in this case 1 plus f plus I will call I will introduce a new parameter fab m7 by m0 into under root T7 by right this is the expression for non-dimensional thrust now if you put fab equal to 0 this is the expression that we had derived for a turbojet without the after burner switched on right when fab goes to 0 this is the expression that we get now we have fab where fab is nothing but m.fuel that is added in the after burner divided by m.air okay so it is the fuel a ratio in the after burner so what do you think of the value of this plus this what is the maximum of f and f plus ab that it can go to it can maximum go to stoichiometric which is what is the value it is 0.067 which is the stoichiometric value you cannot burn more than that so it goes to something like 0.067 now if you go to 0.067 I can still use the condition that f plus fab is less than 1 right with an error of around 6.7% so I will use that I will say f plus fab is very much less than 1 okay so that we can take this terms of and I will also introduce another parameter here just like we had TT4 by T0 as a control parameter we have a new control parameter that is TT6 by T0 right what is the maximum temperature in the cycle divided by T0 the minimum temperature so I will call TT6 by T0 as equal to ?ab remember we had TT4 by T0 as equal to ? burner similar to that I mean introducing a new control parameter to account for the afterburn okay so using this I need still these ratios T7 by T0 and M7 by M0 so I can get T7 by T0 as TT7 into TT6 by sorry this has to be TT6 okay fine these two cancel off and I will get T7 by T0 what is T7 by TT7 it is the ratio of static to stagnation so I can express this in terms of Mach number this would be 1 plus 1 by 1 plus ?-1 by 2 M7 square what is T7 by TT7 by TT6 this is flow through the nozzle right so flow through the nozzle we have assumed the flow to be isentropic so TT the total temperature does not change so this is 1 and TT6 by TT0 we have defined it as ?ab so I get T7 by T0 as equal to ?ab by M7 square now we need to do cascading of pressures to get the other parameter yes yes so why are we doing for the optimal because in case of urbanian optimally expanded if you use a convergent nozzle it is choked we will do both cases okay firstly we will take the optimal expanded flow then we look at what happens if the nozzle is choked the nozzle being choked does not essentially mean that P7 is equal to P0 it can be that P7 is greater than P0 even if the nozzle is choked right so we are taking a special case of choke flow wherein the exit pressure is equal to the ambient pressure now if we do cascading of pressures I get P7 by P0 is equal to okay this is what we get and the first term here is ratio again ratio of static to stagnation pressure this is flow through nozzle flow through jet pipe and then flow through turbine flow through the main combustor flow through compressor and this would be flow through intake and that is ?0 to the power of ? by ?-1 so we will put that down here and we know because we are considering optimally expanded flow what is P7 by P0 this would be equal to unity so we will use that so we can write 1 is equal to 1 by M7 square into flow through nozzles we are assuming all efficiencies to be one then flow through after burner again one then flow through compressor you get ? T then sorry flow through turbine then you have flow through combustor which is one then you have flow through compressor I see and you have flow through the intake which is one and lastly ?0 to the power of ? by ?-1 and here also sorry so I get I can rewrite these two in terms of ? T so I will get 1 is equal to 1 by 1 plus ?-1 by 2 M7 square is equal to ? T ? C ?0 okay and I know from the definition of ?0 that 1 plus ?-1 by 2 M0 square is equal to ?0 so using these two I can get my expression for M7 by M0 I can write M7 as equal to 2 by ?-1 T C ?0-1 and M0 similarly would be and we required in our expression for T7 by T0 this term so I can get that too so I can write T7 by T0 as equal to ? AB by T T C ?0-1 okay so sorry no-1 sorry so this is the expression that we have now we can substitute back and get our expression for the non-dimensional thrust which is S by M. AE0 is equal to or I- I can rewrite this part by taking this out is common I can rewrite this as okay so this is the expression that we get for the non-dimensional thrust now there is still one part missing we know that there is a turbine compressor power balance so these two are not independent variables and they are connected okay so let us get that expression also we had derived that expression for a case without the after burner on do you think it will change with the after burner being switched on because this is something that happens downstream of the turbine it does not get affected and therefore whatever we had derived earlier holds good so from we get Tt is equal to 1- ?0 by ? B into Tc-1 and if you substitute this expression back then we will get a by okay this is the overall expression that we get notice here that ? B is now parameter of very small consequence right the temperature at the end of the combustor is not such an important parameter whereas for the thrust the temperature at the end of the after burner is a more important parameter okay now let us look also at the ISP part okay if we have to calculate ISP we know from our previous expression for non-dimensional specific impulse that I need 1 by F into F by m.e0 so here this was the expression that we had without the after burner how would this change with the after burner I need to include here F plus Feb for the case with the after burner okay so with the after burner on you have this additional term here okay now how do we determine this quantity how do we determine this quantity what do we need to do what did we do to get 1 by F energy balance across the combustor now what do we need to do energy balance across the combustor plus the after burner so let us do that by taking energy after burner we get m.f plus m.fab into Q okay must be equal to m.a into 1 plus F Cp Tt4- Tt3 plus m.a into 1 plus F plus Fab okay this is the portion that gets added on in the after burner into Cp Tt6 by minus Tt5 okay this is the expression that we get now we know that we can use F is very much less than 1 and we also know that F plus Fab is also very much less than 1 so using this I can eliminate these two and therefore I will get F plus Fab that is you take it here and divide this by m.a you will get this into Q by Cp must be equal to Tt6- Tt5 plus Tt4- Tt3 okay now if we look at the TS diagram if we look at the TS diagram this is the TS diagram we have and what we have here is Tt6- Tt5 that is this- this plus Tt4- Tt3 fine now I can add this part what you will get is you will get as though it is a continuous rise from Tt3 to Tt6 okay if I add this part but we know that compressor turbine power balance Cp being the same this portion in this portion is same so you have to subtract one of them if you are adding this you have to subtract this okay and I also know that the Tt2 must be equal to Tt0 right because flow through the what do you have flow through the intake what kind of process is that it is an isentropic process because we are assuming all efficiency is to be one it is an isentropic process total temperature does not change okay so you get Tt2 is equal to Tt0 so we will use that here and do a little bit of jugglery as I said I will add- Tt5- Tt4- okay if you look at this I have not done anything this is what is this part turbine work in the turbine work in the compressor these two must be equal right so I can take this out these two are equal now I know that Tt2 by Tt0 is equal to 1 so this goes to 0 okay so by doing this manipulation I have been able to get this as Tt6- Tt0 which is nothing but you are looking at raising the temperature from here to Tt6 okay this was TS diagram so I can write now F plus FAB Q by CP is equal to Tt6- Tt0 now if I divide by T0 on both sides I get what is Tt6 by T0 this is ?ab and what is this this is ?0 so we get a new expression for we are looking for this expression 1 by F plus F plus FAB is equal to Q by CP T0 ?ab- ?0 okay and following this I can now write isp by a0 is equal to Q by CP T0 ?ab- ?0 into okay so this is our expression for isp with the after burner switched on I will stop here and continue.