 Today's class will start with cycle analysis what we will do in cycle analysis is we look at the Brighton cycle which is mostly followed by all the air breathing propulsion systems and look at what happens to non-dimensional thrust ISP and other parameters if you vary the Mach number okay that is the primary motivation for doing this and we will be able to derive some simple relationships and show where the optimal life value of non-dimensional thrust or specific impulse lies okay. Now firstly we will start with ramjet as it is the simplest system with us with no moving parts and other things so we will start with this and this will be a one-dimensional analysis that is the variation is only along the axis of the ramjet okay so just to refresh we have an intake for the ramjet okay this is the schematic of a ramjet you have air coming in from the left and you have here laded this portion is the intake all this is the combustion chamber this is the CD nozzle now let me call points let me indicate points here I will call this 0 I will call this 2 3 and 4 okay now correspondingly I will put this information on a TS diagram as to what happens to the cycle on a TS diagram let us look at it this is 0 0 to 2 is compression isentropic compression we are assuming ideal processes here 0 to 2 is isentropic compression 2 to 3 is heat addition in the combustion chamber here and 3 to 4 is expansion through the nozzle okay right now we had derived sometime earlier the thrust equation and we had seen that for an air breathing system the thrust is f is equal to m.a into I will call it in terms of the quantity here I will indicate it by v0 and m.a into 1-f v4 that is the exit velocity minus v0 plus the pressure thrust a4 into e4-p0 okay so this is the exit velocity this is the entry velocity and this is the pressure thrust typically the pressure thrust part is very small compared to the convective flux thrust so in this analysis we will neglect this part okay so we will consider that the nozzle is optimal the flow in the nozzle is optimally expanded so p4 will be equal to p0 so this part goes to 0 essentially we are left with only the convective flux okay now we can rewrite this as f is equal to m.a v0 now further I want to simplify it what we know is that velocities are related to Mach number and speed of sound so I will use that information I know v0 is equal to m0 a0 and v4 is equal to m4 into a4 I also know that a0 is nothing but and a4 is so I know the relationship between the velocity in terms of Mach number and then in terms of temperature now we will make an assumption here if you see here a4 corresponds to the velocity of sound at condition 4 at condition 4 you realize that we have added fuel and burned the fuel and therefore the composition of the gases at 4 and 0 are not the same so typically speaking I should have written different notations for these two right but in this analysis we will assume that as the gases pass through the device there is no change to it to either ? or R okay so assuming R and ? to be the same for exhaust and exhaust gases and ambient air then we can do the further simplification of this so I can now write f is equal to m.a I know that v0 is nothing but m0 a0 m4 now if you remember our discussions earlier in the course in all the air breathing propulsion we said that if kerosene is the fuel or any hydrocarbon is the fuel then f the fuel air ratio that is m.f by m.a for stoichiometry is around 6 7 okay so this is pretty much less than 1 so we can neglect f in comparison to 1 okay so f is very much less than 1 so I can rewrite my equation as f is equal to m.a m0 a0 so this goes to 0 I will get m4 into here I know that a4 is ?RT4 and a0 is ?RT0 what I will do is I will take out the ?R part because of this assumption and I will be left with m4 by m0 into under root so we have been able to express this in terms of ratio exit to inlet conditions now what we will do is what is known as cascading of pressures and temperatures we will try to cascade from the nozzle side towards the inlet and see what we can get out of this we will do cascading of pressures and temperatures and see what we can get out of it firstly let me take T4 by T0 T4 by T0 I can write it as T4 by TT4 into TT4 by TT3 into TT3 by TT2 into TT2 by TT0 or I will call this TT0 TT0 by T0 the subscript T here indicates total okay so total or stagnation conditions okay so what we have done is if expressed it is a ratio of static to stagnation and then stagnation of 4 to 3 3 to 2 2 to 0 and then stagnation to static okay so all this will cancel out and you will get T4 by T0 so let us look at what are these values this is nothing but local static to stagnation condition which we can express in terms of Mach number this what does this represent TT4 by TT3 is this by this it represents flow through the nozzle similarly TT3 by TT2 3 and 2 it represents flow through the combustor this represents flow through diffuser or intake this again is local static to stagnation now for an isentropic flow what happens to this ratio TT3 by TT4 if we assume the flow through the nozzle to be isentropic then both these temperatures are same so you get this ratio to be 1 okay then again flow through intake this is also 1 fine your left with now 3 things this flow through combustor and local static to stagnation this the first and the last one we can express it in terms of Mach number okay we will do that right now so this would be 1 by 1 plus ? – 1 by 2 M 4 square okay into 1 and again this would be 1 plus ? – 1 by 2 M0 square okay so lastly we left with flow through the combustor now let me call this ratio TT3 by TT2 as I will call it Tb okay now I can also express that this Tb as TT3 by T0 into fine T0 T0 cancels out you get the same thing this is equal to T0 by okay T0 by TT2 this is flow through the intake which is 1 okay so this is this is 1 so you get this we can express in terms of Mach number okay and let me call TT0 by T0 as ?0 okay so if I call TT0 by T0 as ?0 I will get Tb as equal to TT3 by T0 1 by ?0 I will also call this quantity I will define TT3 by T0 as ?b this is nothing but the ratio of the exit temperature of the combustor to the ambient temperature okay now this is those under the designers control right so therefore there is a very important parameter as we will see later on in the discussions so TT3 is temperature at the exit of the combustor and T0 is the inlet temperature okay so we will call this ?b so I get Tb as this is nothing but ?b by ?0 okay so if I substitute back for this T4 by T0 all this I will get the following expression let me put that as 1 by 1 plus ? – 1 by 2 M4 square 2 1 into if you see here TT3 by TT2 is nothing but Tb which is nothing but ?b by ?0 into flow through intake which is 1 into the last part 1 plus ? – 1 by M0 square I have defined this ratio as ?0 here okay so I will put that down so finally we get T4 by T0 is equal to ?b by 1 plus ? – 1 by 2 M4 square now we have been able to derive this expression for T4 by T0 by cascading temperatures let us now do the same exercise by cascading pressures try and find out what is the relationship for Mach numbers okay so similarly we get P4 by PT4 PT4 by PT3 PT3 by PT2 PT2 by PT0 into PT0 by P0 okay again T indicates the subscript T here indicates total conditions or stagnation conditions. So this is a ratio of pressures static to stagnation the first and the last terms are ratio of start local static to stagnation pressure one is at the exit one is at the inlet okay now this is a flow through diffuser flow through combustor and flow through intake right now this pressure term here the pressure term here this indicates the efficiency across the nozzle okay this indicates the efficiency across the combustor and this indicates the efficiency across the intake so I will call them as this is nozzle efficiency ? and this is burner efficiency B ? B and lastly this is diffuser efficiency ? D okay the other two terms can be expressed in terms of Mach number okay so we get P4 by P0 is equal to 1 by 1 plus ? – 1 by 2 M4 square ? by ? – 1 into ? n into ? B into ? diffuser or intake into 1 by ? – 1 by 2 so this is the expression that we have what is P4 by P0 remember when we started out we made an assumption saying that P4 by P0 is equal to 1 optimally expanded flow. So P4 by P0 is so you get 1 plus ? – 1 by 2 M4 square okay this is the expression that we get by cross multiplying now I will put all these three ratios into one value of ? and I will express let us say ? is equal to ? n into ? B into ? D to the power of ? – 1 by ? if I do it this way then when I plug in for this it will all be a relation of ? by ? – 1 so I can remove the powers and look at what we have here okay so we get finally 1 plus ? – 1 by 2 M4 square must be equal to ? into okay the powers get cancelled out so you are left with this equation just to remind you we had done this exercise earlier the relative values of all these efficiencies ? and typically its flow through the nozzle it varies between 0.96 to 0.98 and ? burner this varies between 94 to 96 and diffuser remember we had indicated this to be a strong function of inlet Mach number if you have a large inlet Mach number you are bound to get lower efficiency because of flow being processed through oblique or normal shocks okay so you will get a lower efficiency if you have high Mach number but as you go down and Mach number you will get a higher efficiency so if you combine the worst case scenario of all this and remember that is a multiplication sign you will end up getting variation ? varies from 0.84 to 0.97 for ? equal to 1.4 okay so the worst case scenario is this 0.84 and the best is 0.97 we will discuss about this when we talk about what is the lowest Mach number in which a ? Z can operate now coming back to this equation here this was our relationship that we were trying to find in terms of Mach numbers what is this term here this is nothing but ? 0 right so we get 1 ? – 1 by 2 M0 square is nothing but ? 0 so from here you get M0 is equal to 2 by ? – 1 ? 0 – 1 and similarly if you put this ? 0 here you can get an expression for M4 as so now we have the relationship for M4 by M0 and T4 by T0 in our earlier thrust relationship we were searching for a relationship for T4 by T0 and M4 by M0 we have got both of them let us plug it in and see what happens so our relationship M4 by M0 is equal to ? ? 0 – 1 divided by ? 0 – 1 and similarly T4 by T0 was ? B to 1 plus okay what is this term this is nothing but this term here right so it is nothing but ? ? 0 so you get ? B is equal to T4 by T0 is equal to ? B by ? ? 0 okay this is because we shown that 1 ? – 1 by 2 M4 square is nothing but ? ? 0 so we have both the ratios that we were looking for our thrust equation becomes F is equal to M0 A0 M0 we had M4 by M0 that is nothing but ? B I firstly put T4 by T0 ? ? 0 into this is the expression that we get for thrust in a ramjet okay now we like to have non-dimensional numbers okay we can easily see that we can non-dimensionalize this thrust if we divide by M. A A0 throughout okay M mark number is not a function of I mean it is a non-dimensional value and all the other things ? B ? and ? 0 are also non-dimensional values okay so I get F by so this is the expression for non-dimensional thrust now we have found out what is the expression for non-dimensional thrust in terms of ? B which is nothing but the ratio of combustor exit temperature to ambient temperature and a function of Mach number and efficiencies okay ? 0 is a function of Mach number and ? is efficiency and M0 is again Mach number itself so the only input parameter that we have under our control is through ? B and the flight Mach number okay the rest of it is system determined that is ? or the efficiency is system determined okay now we have been able to do this we also need another parameter right what is the other parameter of interest to us specific fuel consumption in this case we look at ISP which is 1 over specific fuel consumption is nothing but specific impulse and it is 1 over SFC we define ISP as ISP is nothing but thrust per unit mass flow rate of fuel okay so what we have got is an expression for F by M.A and now I have a definition for ISP like this fine what we have to look at is we have spent a considerable effort trying to derive that expression you have to take a little more benefit out of it so what we will do is we will multiply this suitably and find out if we can use that expression okay so I can write this as F by M.A x A0 x A0 divided by F where F is nothing but fuel a ratio M.F by M.A so if you put M.F by M.A here this M.A and this M.A cancels out and A0 and A0 cancel out you get F by M.F so we have been able to get this expression that we have derived here okay so if you want ISP now the only parameter that we need to derive an expression for is this F here right okay fine we will do that right now we will derive an expression for F by looking at the energy balance across the combustor okay so using energy balance across combustor that is what you are putting in is chemical energy associated with the fuel that is given by M.F that which is the mass flow rate of fuel into the calorific value of the fuel okay you indicates this must be what must this be equal to the sensitive enthalpy rise across the combustor okay so that is given by M.A x the 1 plus here indicates that you are looking at the combined mass flow rate of fuel and air when you are looking at conditions at 3 okay now again using our this is a little weird you are trying to get an expression for F and what we are trying to do is we will try and see if we can eliminate this just to make sure that our algebra is a little more cleaner we will do this F you can compare it with one always in engineering approximations you tend to compare it with a known value and then if it is very small compared to that value you can neglect it okay so you cannot neglect it arbitrarily but you have to make a comparison with some other quantity so if you compare F with respect to 1 we have seen that F is very much smaller to 1 so you can neglect this part here okay so this goes out we get M.F Q is equal to Cp is common for both the burnt gases as well as the incoming air so I can take it out now I need to let me divide both these by T0 okay so I can write this as M.A Cp T0 right I have divided and multiplied by T0 so what is TT3 by T0 when we derived expression for F by M.A what is this expression this is nothing but ? B right and what is this TT2 by T0 is nothing but TT2 by TT0 into TT0 by T0 right what is this ratio this is flow through intake this is 1 and this is what is this ? 0 so you get if you substitute back in here you will get M.A Cp T0 into ? B- ? 0 so finally we can get an expression for F by combining these two we can write F is equal to Cp T0 by Q into ? B- ? 0 and therefore our expression for ISP you see here this expression for ISP that I have if I divide the ISP by A0 I get a non-dimensional quantity what is the unit of ISP ISP is unit is what Newton per kg second right what is Newton Newton is kg meter per second square you get kg by per second so what you will get is meter per second okay so if you split it if you simplify further you will get ISP unit as meter per second a0 is nothing but speed of sound which is also in meters per second so you will get a non-dimensional quantity if you divide it by a0 and if we substitute the expression that we divide for F by M.A0 we will get this as ? B ? ? 0 into ? ? 0- 1 okay this is the expression for ISP by a0 okay we will stop here we will continue in the next class thank you