 In the last few classes we had looked at non-dimensional thrust and ISP for cases wherein the efficiencies were all unity right let us now look at the case where efficiencies are not equal to unity and let us see how the equations change there now if you look at the TS diagram for a turbojet with efficiencies what you will let us consider only the simple cycle no after burner this is a cycle with 100% efficiency now if you have non-unity efficiencies now if you have non-unity efficiencies you will get a cycle like this the dashed ones are for non-unity efficiencies now when we did the analysis for a ramjet we saw that we were able to handle efficiencies of the non-moving parts namely intake then burner then nozzle right here in addition to these we have the compressor and the turbine okay when addition we have good thing about this is we will we are going to look at it in two different ways we are going to handle efficiencies with relate related with the non-moving parts in the same way as we did in the ramjet analysis that is we are going to look at intake burner and nozzle same way as in ramjet here we are going to look at it a little differently okay just to refresh your memory the intake efficiencies can be varying from 0.6 to 0.9 and is a strong function of the Mach number then we have compressor efficiency which can be of the order of 0.8 to 0.87 then the burner is 0.93 to 0.95 then turbine efficiencies and finally nozzle efficiencies of the order of note is one thing that both these two are lower than these two right there is a difference between the two of them that is both these have an adverse pressure gradient to cope with which is why they will have a typically diffuses and compresses will have a lower efficiency compared to turbines and nozzles okay now let us do our cycle analysis cycle analysis the expression for thrust per unit mass flow rate right or non-dimensional thrust what is this expression now this is the same as earlier that is P7-T0 x right this is for optimally expanded flow in the nozzle okay so therefore P7 is equal to P0 that term vanishes out the pressure thrust term vanishes out we have to find expressions for these two ratios okay to find expression for T7 x T0 okay before we go there let us first look at what is different from the all-unit cycle now if you remember in the previous case when we analyze ramjet efficiencies we dealt with slightly differently we said for the non-moving parts if you look at intake burner and nozzle we included the efficiency terms in the pressure cascading the reason for that being that if you assume that it goes to the same same stagnation temperature let us say we are looking at intake on a TS diagram if you are looking at only an intake process this is for 100% efficient cycle now the assumption that we made was a non-isentropic process would go something like this to 2- wherein the temperatures are the same the stagnation temperatures of 2 and 2- are the same only that these pressures are different right these are constant pressure lines these are different and therefore we looked at handling this terms with only efficiencies coming up in the pressure terms fine the flow is at most stagnation it is not stagnation no when at the end of diffuser intake 2 point 2 so 2 we are assuming that it is a stagnation temperature okay but we know as T0 is the content of energy in the flow yes okay if some losses are there so energy should not be equal to the ideal energy like ht0 of the ideal idle is not equal to ht0 dash true so how can we see in real systems what we look at is when we are looking at an intake okay if if the process were isentropic you would have a certain pressure recovery okay at the end of the isentropic compression you would have a certain pressure recovery if the process is real what is the pressure recovery or what is the pressure at the end of such a process is what you are looking at so from that perspective if you look at this here you are trying to capture what is the pressure recovery okay this efficiency would then indicate whether it is 100% efficient 90% 60% efficient the pressure terms contain the efficiency part right and here if you look at this case we have only turbine and compressor to take care of right now in the turbine and compressor what we do is we know that the power of the turbine the work or the power produced by the turbine must be equal to the power consumed by the compressor right because of the balance between the two so if you remember we did not say Tau C and Tau T are two different things we try to combine them through one equation so if we take efficiencies of turbine and compressor there we are going to complete the cycle so even in this analysis we will carry out cascading wherein we will deal with efficiencies in terms of pressure for the non-moving parts and when we come to the compressor and turbine we look at what happens between the power or the compressor and the turbine with efficiencies okay now if you look at compressor turbine power balance or before we go there our efficiency is defined for the compressor and turbine ? C is defined as T T 3- T T 2 T T 3- T T 2 okay and similarly the turbine efficiency is defined as T T 4- T T 5 divided by T T 4- T T 5- okay this is obvious from this figure where this is T T 5- this is T T 3- okay so let us now try and do the cycle analysis and try to get these two ratios so from here if you are going to do a similar analysis to what we did in ramjet okay would whatever we had derived in the previous classes for T 7 by T 0 be any different see compressor turbine power balance will consider efficiency is there but otherwise in the temperatures while cascading we do not look at efficiency true but they appear as tau C and tau T right if you look at the expression they like appear as tau C and tau T when you include that power balance tau C and tau C T cannot be independent terms so when you include them in the power balance that is when you get the real expression for both of connecting both of them okay so let us do the cascading so firstly cascading temperatures okay we want an expression for T 7 by T 0 similar to the previous cases T 1 by T T 7 into T T 7 by T T 6 T T 6 by T T 5 T T 5 by T T 4 T T 4 by T T 3 T 0 okay. Now as in the previous case what is this this is a ratio of stagnation to static so I can express it in terms of Mach number this is flow through jet pipe this is one now this is flow through nozzle one this is flow through jet pipe again one this is pi T sorry tau T and this is flow through the main combustor one and this is tau C this is one flow through intake and this is ? 0 right so it is similar to what we had earlier derived that is sorry this has to be T T 4 by T B and then if we put it in terms of ? B the expression that we get is we know that T B and ? B are related that is so from here I get T B would be equal to ? B by T C ? 0 and if I substitute it finally I will get T 7 by T 0 is equal to T T T C this and this cancels out I get okay now I need an expression for Mach number ratio which I will get by cascading pressures so we know that P 7 by P 0 is equal to 1 equal to P 7 by P T 7 P T 7 by P T 6 x P T 6 by P T 5 P T 5 by P T 4 now when we cascade pressures if you remember what we did with ramjet we will get efficiencies here so this is one is equal to 1 by 1 plus ? – 1 by 2 M 7 square x P T 7 by P T 6 this is flow through nozzle so this has an efficiency that is the nozzle efficiency okay now flow through the jet pipe also as an efficiency let me call it ? JP if it is on it will be different value if it is off be a slightly different value then what is this P T 5 by P T 4 this is ? T x ? burner this is through the combustor then this is through the compressor so this is ? C this is intake x ? 0 to the power of now just like when we did ramjet analysis we will club all these efficiencies into one and we will define ? ? as if I define it this way I can express all the terms here as a power of ? by ? – 1 okay this is as the intelligent way of putting it here so that we will get the same powers so ? T I know is nothing but ? T to the power of ? by ? – 1 I will put all of them so I will get 1 plus M 7 ? C ? 0 okay if you look at this expression we wanted in the denominator that value so we have got this so now I can write T 7 by T 0 as equal to I can cancel the ? T terms write it as ? 0 now we are interested in mark number ratios and the other thing that I know about us 1 plus ? – 1 by 2 M 0 square is equal to ? 0 so using these two expressions I get the mark number ratio as M 7 by M 0 is equal to ? T T C ? 0 – 1 okay now we know both the ratios temperature as well as Mach number so we substitute them in the equation and find out how it looks like so when we put them in this non-dimensional thrust equation M ? a 0 we get ? B by ? if we put efficiencies is equal to 1 here we recover back the turbojet equation right so we have put this is equal to 1 which is what is the turbojet equation right now we need to find an expression connecting Tau C and Tau T right so this comes from from compressor turbine power balance we get M ? a Cp T T 3 – T T 2 this is the actual power consumed by the compressor this must be equal to M ? a Cp T T 4 – T T 5 dash this is the actual power delivered by the turbine okay. So here we have assumed Cp to be constant yes you can take a mechanical efficiency which will turn out to be that you will have to take one by mechanical efficiency here you can include that you can assume it to be 1 if you want to include it you can put it here okay. So Cp is equal to constant and the other assumption is that I am sorry to write here 1-f because there is a larger mass flow rate and the other assumption that we will make is f is very much less than 1 okay now using this we will get we also know how to connect efficiency of the compressor to this right you remember what was the efficiency T T 3 by – T T 2 divided by okay so I get I want this term I want this term sorry so I will get it as T T 3 – T T 2 divided by ? C and similarly turbine efficiency was T T 4 – T T 5 dash divided by okay. So if I take these efficiency terms into account these equations will we will be able to rewrite as follows okay right I think this mechanical efficiency needs to be included here and not here because if mechanical efficiencies are non-unity which is less than 1 then the compressor requires more than what the turbine can give so you have to include it here so it will be if you cross model reply it will be a fraction compressor will only get a fraction of what the turbine develops okay fine so where does mechanical efficiency come in here in this equation right fine so we are interested in T C and T T so to get that we divide both sides by T 0 T T 4 by T 0 – T T 5 by T 0 what is T T 3 by T 0 T T 3 by T T 2 x T T 2 by T 0 right what is this T T 2 this is 1 for flow through diffuser or intake this is what is this now see and this is ? 0 so this term becomes ? 0 and similarly on this side what is T T 4 by T 0 this is ? B and this term would be T T 5 by T T 4 x T T 4 by T 0 okay so T T 4 by T 0 is ? B this is so T T x ? B so we get ? 0 by ? B x ? C ? T mechanical right or T T is equal to 1 – ? 0 by ? B now if you look at this expression what happens if efficiencies are not equal to 1 then this term increases right and you will get a lesser T T fine that means that you will have lower and lower pressure at the end of the turbine fine that correct sorry if you have unity you will have ? 0 by ? B x T C – 1 but if you have non unity then what happens this term is large and therefore 1 – this will be small which is again what I said earlier that is correct so you will get T T to be small fine and then T T is small the pressure ratio the pressure at the end of turbine is low and sometimes if the efficiencies are very poor one might actually end up having a turbine at the end of which there is no power left for there is nothing left for expansion through the nozzle which means there will be no thrust that is produced so one needs to be careful about efficiencies here so that we do not end up in such a situation okay now we have been able to derive expressions for non-dimensional thrust for all situations that is efficiency is not equal to one efficiency is equal to one and optimally expanded flow as well as nozzle being choked and we have also looked at after burner without after burner and what is the other mode of water methanol injection also we looked at right now here in this case ISP by a0 that does not change the expression does not change but f by m dot a0 will be different because you now have efficiency terms coming okay the expression for ISP by a0 would be now I sorry I think we need to derive this see there is efficiency term involved and therefore the TT4 by because of which we need to re look at what is the ISP expression so if we look at ISP by a0 we had in the previous classes looked at this expression this is 1 by f into f by m dot a a0 okay now how do we get 1 by f we get 1 by f by looking at the energy balance across the burner so from across the burner I get m dot f into Q is equal to m dot a into 1 plus f remember in the earlier expression that we wrote when the efficiencies were 1 we were looking at TT3 only right because if you look at what happens due to non-isentropic process this temperature would be higher than or different than the earlier temperature so you have a situation where in this is different so therefore that will come here right so we make the assumption that we done earlier that is f is very much less than 1 so I can take out this quantity and I get my expression for 1 by f is equal to TT4 by T0 now we know that compressor efficiency is ? C is equal to TT3 – TT2 okay so using this goes to the denominator so it will be we know efficiency is denoted in this fashion so I can connect from TT3 – to TT3 here so 1 by f is equal to Q by CPT0 before I do that I will just do a small manipulation with this expression TT3 by TT2 – 1 into TT2 I can write this expression rewrite this as this and TT3 – by TT2 – 1 this and this cancels out I am interested in this expression so I will get TT3 – by TT2 as equal to TT3 by TT3 – by TT2 would be 1 – what is this TT3 by TT2 this is tau C okay so 1 plus this goes here you get this now we have been able to get the expression for TT3 – by TT2 we can use it here and we can write TT3 – by T0 as equal to this expression is what we have derived just now what is this this is nothing but ? 0 so I get TT3 by T0 is equal to ? C – 1 divided by ? C into ? 0 so finally my 1 by f term comes out to be ? B – ? 0 to 1 plus tau C – 1 okay this is the expression that we get so what happens to if you plug this in in the ISP by A0 expression you get ISP by A0 is equal to Q by CP T0 is the expression that we get for ISP by A0 now if you see in this expression what happens if ? C is not equal to 1 this part reduce increases and therefore ? B – this would be small right so can I now say it is advantages to have a non-efficient compression process if you look at this expression it is a it is in a sense it disguises some things we have not written the full expression for F by M. A0 where in efficiency terms are also there right if you look at this it appears as if this part will be reduced will be increased and therefore the entire term will be smaller and as a consequence ISP will be higher but if ? C becomes large that means is is a low value not large it is a low value then what happens is the pressure ratio across the compressor or the pressure at the end of the turbine becomes smaller and smaller they will you will not have any power left for expansion through the nozzle right so although it looks very deceiving here it is not the actual case you need to plug in the values for ? C and other things into this expression and then see how it looks like we look at it in the next class thank you.