 In the last class we had looked at the cycle analysis for a ramjet and we had started the cycle analysis for a turbojet let us continue the cycle analysis for the turbojet if you look at the TS diagram firstly let us assume that all processes are 100% efficient that is we will assume an ideal cycle initially and we have also started something wherein the flow is optimally expanded in the nozzle these were the two things that we are going to look at if we take an ideal cycle then there is compression in the intake itself so I will call 0 to 2 and then through the compressor then you have heat addition then you have expansion through the turbine then you have expansion again through the nozzle this is the TS diagram we have okay now we had derived certain things in the last class we had derived that ? B is nothing but ? B by ? C ? 0 okay and we got this expression for F by m dot a a 0 this is the expression that we had okay now let us try and find out how to get these ratios just like the previous time wherein we had done this cascading in ramjets let us do the cascading and find out how we get these temperatures now to get T7 by T0 can you cascade T7 T7 by TT7 into TT7 by TT6 by TT5 TT5 by TT4 TT4 by TT3 TT3 by okay again the subscript T here indicates stagnation conditions and if you take a look at this all these cancels out and you get TT7 by T0 okay so we also know that the last term is nothing but ? 0 from our previous class so we have this is nothing but a ratio of stagnation to static so we can express this in terms of Mach number as 1 plus ? – 1 by 2 M7 square okay now what is TT7 by TT6 this is flow through the nozzle again if the efficiency is a 1 the stagnation temperature ratio will be the same so this is 1 and TT6 by TT5 this is flow through the after burner anyway we are not considering in this analysis this analysis is without the after burner so this will again be 1 and TT5 by TT4 is flow through the turbine now flow through the turbine because it is again an isentropic process this ratio will also be what will that ratio be TT5 by TT4 you have forgotten something we derived in the last class that this is ? T and TT4 by TT3 is process through the combustor so I will call it ? B into TT3 by TT2 is flow through the compressor again this is ? C into again you get the diffuser diffuser this ratio because the flow is isentropic is 1 and into ? 0 okay so we know that ? B is this if we substitute it there we will get T7 by T0 is equal to ? T into ? B is ? B by ? C ? 0 into ? C ? 0 this is the ratio that we have this ? C ? 0 and this ? C ? 0 cancels off and we get ? B T T divided by 1 plus ? – 1 by 2 M7 square okay now similarly we will do for the pressures P7 by P0 is equal to now we assume all efficiencies to be 1 so the first one this would be if you cascade it it will be P7 by PT7 into PT7 by PT6 PT6 by PT5 okay this is the cascading and if we plug in the values then this is the ratio of static to stagnation condition so you get 1 plus ? – 1 by 2 M7 square to the power of ? by ? – 1 then this is flow through nozzle so you get for all efficiencies being 1 you get 1 then you have flow through after burner is again 1 what is this flow through turbine you get ? T okay pressure ratio across turbine and what happens in an ideal cycle to the pressure ratio across the combustor for an ideal cycle the pressure ratio across the combustor it will be isentropic isobaric process so the pressure is the same so this ratio would be 1 and PT3 by PT2 this is again ? C ratio pressure ratio across the compressor and PT2 by PT0 is flow through diffuser for an ideal process this is 1 and lastly this is nothing but ? 0 to the power of ? by ? – 1 okay fine now I can change this to ? T and we will get a new ratio so I will do that I will get ? T ? C ? 0 divided by 1 plus okay this is the ratio that we get and if you remember we have taken the case where the flow is optimally expanded through the nozzle so what is p7 by p0 1 so we can rewrite this as this ratio as 1 plus ? – 1 by 2 m7 square must be equal to ? T ? C ? 0 right I can take out the powers without any problem and therefore I can write it like this now remember in this expression for T7 by T0 in the denominator I have 1 plus ? – 1 by 2 m7 square which is what I have got there so if I plug in this value I will get T7 by T0 is equal to ? B ? T divided by ? T ? C ? 0 so I will get this is the temperature ratio now to find the Mach number ratio we will have to take this expression and derive the ratio for Mach numbers so we know that 1 plus ? – 1 by 2 m7 square is equal to ? T ? C ? 0 and we also know 1 plus ? – 1 by 2 m0 square is equal to ? 0 so from these two expressions I can write m7 is equal to ? C ? T ? 0 – 1 and similarly m0 we had done this earlier is equal to okay and the ratio m7 by m0 will then become so this is the ratio of Mach numbers and that is the ratio of temperatures that we were looking for we got those two and if we plug them into the expression for thrust f by m.a0 okay that is what we were looking for so f by m.a0 is equal to what we got was m0 x T7 by T0 x m7 by m0 – 1 if we substitute the ratios for T7 by T0 and m7 by m0 we will get okay this is the expression for non-dimensional thrust that we were looking for now is this expression complete is there something that is still missing see here you have got ? C and ? T right but we know that compressor and turbine power balance must be there so they cannot be independent of each other they must be related to each other and therefore you can write one expression for connecting these right so there is a relationship for connecting T C and T T that is through the compressor turbine power balance now if we do the compressor turbine power balance we will get this ratio connecting T C to T T now we know that flow through compressor is m.a Cp T T3 – T T2 this must be equal to m.a x 1-f T T4 – T T5 now there are two things that we are going to make an assumption on if you let take a look at this this is primarily air getting compressed in the compressor whereas for the flow through the turbine you have product gases if you remember earlier when we said ? and R are the same for both air and exhaust gases right if ? and R are the same for both air and exhaust gases then Cp what should happen to Cp Cp should also be the same for both of them so assuming Cp to be constant and also we will make another assumption that we have been doing all along that the fuel a ratio f is very much less than 1 when for a turbojet because typically this will be of the order of 0.02 to 0.04 so even if you neglect it you are going to make a 2 or 4% error okay so with these two assumptions this will simplify to T T3 – T T2 must be equal to T T4 – T T5 so what we do from here is we know by cascading what the ratios are so we just have to divide both sides by T0 let us divide both sides by T0 okay now what is TT3 by T0 again if we cascade I will get TT3 by TT2 x TT2 by TT0 x TT0 by T0 what is TT3 by TT2 this is flow through compressor so this will be Tc and this ratio is 1 across the diffuser so and this is ?0 right TT0 by T0 is ?0 and this would be TT2 by TT0 x TT0 by T0 this again is TT2 by TT0 is 1 and this would be ?0 fine and what is T4 by T0 this is ? B this is from our definition and TT5 by TT0 I can write it as TT5 by TT4 x TT4 by T0 what is this TT5 by TT4 is ? T this is ? T x ? B okay so I end up getting if I take out ? 0 as common ? 0 x ? C-1 must be equal to ? B x 1- ? T so therefore I can write the expression for ? T as I can write ? T as equal to okay so this is the expression for T T now if I plug in this expression for T T in this expression for F by M.A0 that is the non-dimensional thrust before we do that what we see here is we need this ratio all the T T and T C terms are only in this ratio so let us look at this first term in the under the square root sign and let us try to simplify that so T C ? B ? 0 divided by T C ? 0- I can write the numerator combining these terms these are the terms containing T C and T T so if I combine these terms and rewrite it I will get it like this okay which is I can simplify this further and write it as T C ? 0 cancels off here I will get ? B T T- ? B by T C ? 0 and now if I substitute for T T in this expression okay I will get this as equal to ? B- ? 0 T C so that is now if I plug back this expression into that and rewrite my expression I will get the non-dimensional thrust as F by M.A A0 must be equal to this is the final expression for a non-dimensional thrust. T T and T C are connected know T ? B is a controlling factor is okay with you because it is a turbine in let temperature so that is fine with you what you are looking for is why is it that T T has gone out the T T had to go out because the compressor and turbine there is a power balance between them and we have assumed the processes to be at 100% efficient so the compressor power must be equal to the turbine power. So when you do that you can eliminate T T and that is what we have done here so it will be only dependent on T C ? 0 and T B now we have done all these calculations right what we need to do is cross check whether what we have got is correct how do we do that how do we cross check what you got is correct or not still does not tell us anything what do we know you will always assume previous class is true right or previous class is correct so we know the results for ramjet right what is the result for ramjet the result for ramjet was f by m.a a 0 is equal to m 0 ? B by ? 0-1 for ? is equal to 1 this is for okay this was the result that we had for ramjet now what is the difference between ramjet and a turbojet that you have a compressor right compressor and therefore a turbine so what happens if you put T C to be equal to 1 compressor pressure ratio or ? C is 1 or T C is 1 compressor turbine temperature ratio is 1 or compressor pressure ratio is 1 that is what is a ramjet right so if you put T C is equal to 1 here what happens ? 0 ? 0 cancels off and you get T C this is ? B by ? 0 right so let us do that turbojet becomes a ramjet when T C is equal to 1 so when I substitute T C is equal to 1 in the expression for ramjet turbojet you get ? B- ? 0 into 1- ? 0- ? B by ? 0 T C is again 1 okay. Now it is obvious that you can cancel out this and you get you can take out ? B common so you get m 0 and if you take out ? B common what you get is 1-1 by ? 0 okay this is anyway the same as you you can take 1 by ? 0 out you will again get so I get which is the same as ramjet so in that sense it is consistent okay what we have derived is consistent with when we put T C is equal to 1 we get the ramjet result okay now there is another thing that we need to look at what is that we know that turbojets have a static thrust right turbojets do produce a static thrust now we need to find out whether the expressions that we have derived do show that now in this expression here what happens what is the condition for static thrust if you put m 0 is equal to 0 then you get 0 no suddenly we have done all this extravagant calculations and suddenly found out that we are on the wrong side our equations do not bring out that fact that it still can produce you know static thrust looks like it produces 0 thrust or is there a catch to it let us see see we know that what is ? 0 1- ?-1 by 2 m 0 square okay so if you you can you have m 0 and ? 0-1 so m 0 by under root ? 0-1 is equal to what do you get here under root right so we can use that here and rewrite the expression let us do that so I get okay is this correct right we have substituted for this in the earlier equation and now using this if you put here m 0 equal to 0 this goes to 0 so the expression for static thrust would be static thrust of a turbojet so you get okay so this is a non-zero quantity so therefore we can safely say that whatever we have derived is correct and it produces non-zero static thrust okay even when m 0 is equal to 0 what happens when m 0 is equal to 0 what happens to ? 0 ? 0 sorry when m 0 is equal to 0 it means that ? 0 should be 1 so you substitute it here again you can rewrite this expression ? 0 so this can not go to 0 and therefore we have a positive static thrust okay and the static thrust itself is some air moving in the compressor itself some air was sucked in the compressor so it got some velocity yes so is it I am not really 0 what is the takeoff velocity of an aircraft okay let us do that see so you have takeoff velocity around 250 km per hour what is the speed of sound at that condition meters per second so you convert it into kilometers per second kilometers per hour 11 so I will assume it to be fine around 1200 km per hour so you calculate the Mach number based on this what will be so what is this sacrosan about 0.3 Mach number which we say is the regime where we differentiate that the flow is incompressible and then the flow becomes compressible what is this limit of 0.3 so if you look at this number here Mach number is 0.21 so if we do 1 plus ? – 1 by 2 this is m0 square is nothing but ? 0 how different will it be from 1 this will be anyway close to 1 know so even while the aircraft is taking off with very high velocities of around 240 250 km per hour the Mach number is around 0.2 to 0.3 which means that it is still ? 0 is still around 1 okay so therefore this is consistent what we have done here is consistent right okay now let us look at what happens to the other quantity of interest to us that is ISP okay in the previous class when we were looking at ramjet we had already done this exercise and we noted that ISP by a0 which is the speed of sound we can write this as 1 by f into f by m.a e0 right and we said that we put in a lot of effort to find out this expression so it is meaningful to use this to get the ISP so we know this part and we need to evaluate what is this quantity right so let us do that or we can evaluate this by again looking at the power balance in the combustor from energy balance across combustor I can write m.f q is equal to m.a x 1-f okay fine and what we will again do is we are trying to find an expression for f we will say that f here in comparison to 1 a small so therefore we will neglect that we can do that even while trying to derive this expression for f so f is nothing but m.f by m.a so using this so I can rewrite my expression as 1 by f must be equal to q by CP I will take out T0 as common then I will be left with TT4 by by T0 okay what is TT4 by T0 this is ? B what is this TT3 by TT2 x TT2 by TT0 T0 this is ?0 this is 1 this is ? C so you get 1 by f is equal to q by CP T0 x ? B- ? C ? 0 okay so this is the expression into now we have got 1 by f there so I can write the complete expression for ISP by a 0 as equal to ? 0 T0 C plus ? 0 so this is the expression that we have for ISP by a 0 we will stop here and we will continue in the next class thank you.