 In the last class we had derived expressions for ISP and non-dimensional thrust for a case with the afterburner switched on and for the flow through the nozzle we had assumed that the flow is optimally expanded through the nozzle. Now let us take the other case wherein the flow through the nozzle is a choke flow okay and efficiencies are all 1 and we will see how this case looks like so we had done a very similar analysis without the afterburner being switched on and we had got this expression for non-dimensional thrust as f by m.aa0 this was the expression that we had for the non-dimensional thrust for a choke flow through the nozzle with all efficiencies being 1 and this is the same expression that we will have with the afterburner switched on also only these quantities change okay. So for afterburner switched on we need T7 by T0 and in the previous class we had seen that this is nothing but ?ab by 1 plus ?-1 by 2 m7 square right we know that for a choke flow through the nozzle what is it that we know m7 must be equal to 1 so substituting it here we will get that it this is the expression for T7 by T0 and similarly we need expression for P7 by P0 when we cascade we will get this 1 plus ?-1 by 2 m7 square right into ?t ?c okay this is the expression that we get now I also know that I can write ?t as ?t to the power of ? by right and I will substitute these because all of them will then be raised to the same power so it will be easier for me to handle that I will get and I also know that m7 is equal to 1 so substituting these I will get okay so we have got pressure ratios also then for getting we have got this ratio we need to get this quantity okay just like in the previous case we will get this by looking at mass flow rate through the nozzle considering mass flow rate through the nozzle we get that m.a into 1 plus f plus fab in this case must be equal to ?7 v7 a7 now again we can use what we have done earlier that f plus fab is very much less than 1 then I will get m.a is equal to ?7 I can write it as P7 by RT7 into v7 is a7 into m7 okay this I know is 1 okay for the choke nozzle so a7 if I write it as ?RT7 I will get P7 into under root ?p7 divided by now again we multiply both sides by a0 okay so I get what I was looking for m.a0 is equal to P7 under root ?a0 is again ?RT0 so I will use that so I can cancel out R and R here and I will be left with ?P7 a7T7 now if I divide both sides by P0 by P0 a7 you get the expression that we wanted m.a0 divided by P0 a7 is equal to ? into P7 by P0 and we know the values of both of them P7 by P0 and T0 by T7 so if we substitute it we will get this is the expression that we get now we know all the three quantities that we were looking for so I can write f by m.a0 as equal to under root T7 by T0 which is nothing but 2 ?ab by – m0 plus P0 a7 by m.a0 that is 1 by ? into P7 by P0 – 1 that is the same quantity okay this is the expression that we get for f by m.a0 for an choke flow condition with the afterburner switched on you can also similarly get an expression for ISP only thing that needs to be changed is that f by m.a0 is this expression right all the other things in the ISP by a0 expression that we derived earlier was good so you just need to change the f by m.a0 expression so you have now got both things that is for an optimally expanded flow and for a choke nozzle choke nozzle is the more general case and optimally expanded flow as I said is a very special case of choke nozzle wherein the exit pressure is equal to the ambient pressure okay now just like what we had done earlier that is look at typical calculations and see what happens with the afterburner turned on okay. I will use the same set of numbers that I had used for a choked flow in the nozzle so we will keep the flow in the nozzle as choked and use the same set of parameters and find out what happens with f by m.a0 and ISP with the afterburner turned on so first is takeoff that is m0 is equal to 0 and at 11 kilometer altitude m0 being equal to 0.8 so we look at firstly f by m.a0 and then we look at ISP SFC, SFC is in kg R and we look at two ratios that is okay and we look at two conditions here one no AB that is no afterburner and then the other one with afterburner again here same thing now we are using all these parameters the same so I will get f by m.a0 as 2.27 and 2.15 similar to the last case that we had looked at now let us look yes both cases being for the choked nozzle so let us look at what happens now with the afterburner being switched on all this is for a choked nozzle now let us look at what happens to these values when we switch on the afterburner with afterburner this goes to a fairly large value of 6.07 sorry 3.67 and ISP reduces to 29.5 and this increases to 1.2 and similarly this goes to 3.85 1.37 and this ratio of f by m.a0 that is the non-dimensional thrust how much do we get an increase from with the afterburner being switched on to without it this ratio would be 1.62 for this case and 1.79 or 1.8 for the second case okay and SFC ratio is what we need to pay to get this kind of ratio if you look at this okay what we see is that we need to really shell out a lot more fuel if we have to get a thrust increment by this much if we have to get in this case 60 to 80% thrust increase we have to spend something like 50 to 80% on the SFC right so it is a very large SFC increase compared to what you get for the thrust you can also put out the other way that if the same thing was done let us say if we had if you remember our discussions wherein we talked about the limitation on turbine inlet temperatures we said turbine let temperatures cannot be raised because of material considerations and therefore we limited by using excess air now suppose we were free to increase this to a larger value then the same kind of thrust increase remember turbine inlet temperature is in the hands you can give the more fuel to the main burner or less fuel but there is an overall limit right that can be changed within certain limits if we were allowed to increase the turbine inlet temperature the same kind of what you see as f by m dot a ratio could have been obtained with same can be obtained with same ratio can be obtained with 35% if turbine inlet temperatures could have been increased okay. So what we are saying is you are spending 80% increase in SFC while the same thrust increment could have been got if you were allowed to increase the turbine inlet temperature this shows the need for you know better materials wherein we can go for a larger turbine inlet temperature and wherein you can say one fuel if you want excess thrust you can increase the turbine inlet temperature and get the same excess thrust at a much lower cost why does this happen why do we say that if we increase the turbine inlet temperature if we are allowed to increase the turbine inlet temperature the SFC would reduce whereas if you switch on the after burner you see that SFC increase for the same thrust increment is more why does this happen at those velocities that is the one that gives the higher thrust that is not the reason what I am asking you is if you look at the two cases SFC increase with the after burner switched on is much more for the same thrust increment than the case if we take wherein we are allowed to increase the turbine inlet temperature turbine inlet temperature you could have got the same with a 35% increase whereas you are spending around 80% increase in SFC to account for the same thrust increment why should this happen as my question why should burning fuel in the after burner be more expensive now if you go back to our discussions wherein we talked about the need for the after burner I talked about something known as availability right if you add heat at a very high pressure then you have a opportunity to expand for more than if you add heat at a lower much lower pressure what we are doing in the after burner is we are we have passed the turbine where the flow has expanded already and the pressures are very low and there we are trying to add heat which means that the opportunity to expand is smaller and therefore we find that the SFC will increase because you are now expanding from a lower pressure to even lower pressure and but you are spending more fuel to increase that temperature okay now let us look at the next method of getting thrust augmentation that we discussed earlier that is water methanol injection okay now if you remember earlier discussions I had said that you can add water in the main combustor and that will because the flow at the exit of the combustor is choked it will act as though it is increasing the pressure inside the combustor okay and we use water for that now water at very low temperatures that are encountered at higher altitudes would tend to freeze therefore we need to add some additives to make sure that it does not freeze that is why methanol is added and also methanol provides the additional heat that is required to increase the temperature of water from ambient to the turbine inlet temperatures okay so let us look at how this system works so we are going to consider the case where water and methanol are injected in the main combustor okay now what is our non-dimensional thrust equation that is optimally expanded flow and ? being 1 okay so this is the expression that we had for non-dimensional thrust for these conditions right now if we are injecting water and methanol in the main combustor I need to add a term here okay it is not just 1-f it should be f plus I will call this fw whereas where fw is nothing but m dot mass flow rate of water plus methanol divided by mass flow rate of air okay so this is the expression that we have now here in this case I cannot say that combining these two it is much less than 1 because we will see that you can increase this to something like 30% of the overall flow so I cannot neglect this part compared to 1 so I have to retain it and do all the rest of the algebra again we need expressions for T7 by T0 and M7 by M0 okay now in this case if you look at the expressions that we have derived for M7 by M0 okay we had got earlier M7 by M0 as equal to ?0-1 and T7 by T0 as equal to ?b by ?c ?0 right both these expressions do not change the only change in this case comes about when we are looking at the compressor turbine power balance if you are adding water methanol in the combustor all the other expressions these expressions remain the same the only change will come in the compressor turbine we look at how that happens so I will say no changes here to the case without water methanol injection okay so let us look at the compressor turbine power balance okay now as I said if you have if you add water in the combustor there are two things that happens one is the flow is choked at the exit of the combustor if the flow is choked then we know that it is only a function of upstream conditions upstream you are increasingly increasing the mass flow rate right you are adding water and methanol you are increasing the mass flow rate which means that the pressure upstream will also have to increase so what it essentially does is increases the pressure ratio across the turbine okay it acts as though the compressor pressure ratio is higher fine so this combustion chamber pressure increases and therefore we find that it acts as though the compressor pressure ratio is increased now earlier we had a compressor pressure ratio of Pisie okay now which was nothing but PT 3 by PT 2 now because of the addition of water and methanol in the main combustor this will get changed to this will become I see not into 1 plus F plus FW divided by 1 plus F that is this is the new compressor pressure ratio it is the earlier compressed pressure ratio excuse me it is the earlier compressor pressure ratio that you have multiplied by the increase mass flow rate divided by the earlier mass flow rate so there is a new compressor pressure ratio that you will get so if you look at the equation for the turbine compressor power balance what you will get is m.a Cp because this pressure ratio across the compressor has changed it acts as though the temp the pressure ratio you will find that you can also express this as in terms of Tc0 right and if we do the analysis if we do this analysis what we will get is the expression for Tt as equal to earlier it was 1- ?0 by ?b x Tc-1 this was the earlier expression now there is an additional mass flow rate that is going through if you remember we had neglected 1 plus F which would be in the denominator now there is an additional mass flow rate that is going through and that cannot be neglected so you will have 1 plus F plus FW here okay and the Tc part ?c was earlier different now the ?c would be ?c would be equal to Tc0 into 1 plus F plus FW divided by 1 plus F okay so this is the expression that you will have for Tc and if you plug it back in you will get the new expression for Tt so you will get finally Tt is equal to 1- ?0 by ?b into Tc0 1 plus F plus FW divided by 1 plus F so what really happens is the actual value of Tt because you have increased the mass flow rate and because you have increased the pressure upstream of the turbine both the pressure and the temperature at the end of the turbine would be higher then compared to without water methanol injection which means that if you expand this flow through the nozzle you have more scope for expansion as well as the temperature at the end of expansion would be much higher therefore you get a larger thrust now obviously you have to pay for it so ISP has got to be higher in this case why should ISP be higher yes water we are injecting as a spray this needs to be water which is a T0 needs to be increased to the turbine inlet temperature so you need to obviously burn fuel to do this yes but still the latent heat you have to provide for now at least to bring it up to TT3 you need to provide for it so this will mean that there is an increase in ISP so if you take a look at ISP by a not expression it will change to Q by CP T0 into earlier we were neglecting this part 1 plus F plus FW we have to have earlier we did not have this FW and we said F is very much smaller so we neglected it now you have that part into ? B- ? C ? 0 plus L prime into FW into F by okay now L prime is nothing but L by CP T0 where L is heat absorbed by water to rise its temperature from a liquid at T0 to gas at TT3 then the remaining part is accounted for in this portion okay this is where you are taking it from TT3 to the turbine inlet temperature TT4 this part accounts for the latent heat of water that you need to supply for raising its temperature from T0 to turbine inlet temperature okay. I have to say here in this analysis we have looked at only water you not accounted for you can account for it by suitably changing the L value okay L value in that case will get reduced so the ISP part will be a little higher okay now if you were to do the same analysis as we had done earlier that is make a table and look at what happens with increasing percentage of methanol injection how much is the thrust increment that we get and what is the cost that we need to pay we will use the same set of values that we use for the choke nozzle that is ? B is equal to 5.55 this is at 11 kilometer and M0 is equal to 0.8 tau C0 would be 2.03 and therefore ?0 would be 1.128 and F let me keep it as 0.03 E0 is equal to 294 meters per second if you have this then for varying fractions of FW okay we can get tau C as I said only that change are in that expression are tau C and tau T so you have 0 then it is 2.03 and this is 12 and this would be 0.8 2.46 which was what we had seen earlier now I can put this in kN so I will get 41.8 now if we increase it to something like 0.2% then compressor pressure ratio will rise and correspondingly the temperature ratio will rise it lacked as though the compressor is giving out air at 14.5 bar I mean 14.5 pressure ratio whereas the compressor was giving out at 12 it acts as though it is giving out at 14.5 and correspondingly this will also increase and you get a higher thrust but your ISP will be reduced and if you further increase it to something like 0.3 this goes to 2.18 and the compressor pressure ratio increases this increases also so we see that as the F by M.A0 increases we are also getting a decrease in ISP this is because we need to add more heat here to increase the temperature okay then I will stop here in the next class we look at what happens if we have efficiencies to deal with in the turbojet okay thank you.