 Good morning in this class we will discuss what is the role of efficiency that is repulsive efficiency as well as well as the overall efficiency for both air breathing and non-air breathing. But before we go there as if wanted to ask you this question let us say we use a rocket motor inside water for a torpedo is it going to be useful or do you think it is not going to be very effective what is your opinion you understand for an underwater torpedo if you are looking at using a rocket motor torpedoes usually have a propeller driven system if we use a torpedo with a rocket engine instead do you think it is going to be very effective or not going to be effective which are not effective okay let us see as we go along in this class what is the reasons and what whether it is going to be effective or not okay. Now firstly just to recollect that if you are looking at a aircraft engine you can consider it as a box the air breathing engine you can consider it as a box wherein you are inputting certain mass flow rate of fuel and certain mass flow rate of air comes in through the inlet at m.a va and p a and this goes out the mass flow rate is m.a plus m.f and V e be the velocity and let P e be the pressure here and ambient pressure be p a now we know that the thrust of this everything engine would be m.a into 1 plus f V e – V a now f is nothing but ratio of mass flow rate of fuel to mass flow rate of air typically for aircraft engines we have seen this for stoichiometric it is around 0.067 but usually we operate it less than that so it is around going to be around 0.02 to 0.03 so f is we can make an assumption that f is very much less than 1 and therefore neglect this part neglecting this part the error is typically of the order of 2 to 3% only okay so if we neglect this part and if we also neglect the pressure thrust component which is usually small in aircraft engines so neglecting pressure thrust and using f being very much less than 1 we get f is equal to m.a V e – V a as the thrust equation this is the thrust equation that we are going to use in this class for all the further calculations that we do okay. And similarly for rockets for non-air breathing engines we can assume this is a box right where the pressure at this point is P e ambient pressure is P a and m. is the mass flow rate and V e is the velocity remember rocket motor or a non-air breathing engine carries its own fuel and oxidizer therefore it does not depend on air flow through it okay and there is no intake asset okay all the mass that is needed to be thrown out is generated inside the rocket motor itself. So here again thrust we had derived this expression before as m. V e plus a e P e – P a right now again if we assume that the contribution of pressure thrust is small then we can write thrust for a rocket motor as m. V e okay now notice that V e is very similar to some other quantity that we had defined earlier called IS right so we get this expression for thrust and this is the expression that we are going to use in this class for rocket motors this is the thrust and for air breathing engines this is the thrust that we are going to use okay fine. Now if you look at let us say we have this much of energy in a air breathing engine as well as in a non-air breathing engine what we have is chemical energy stored in the propellant of the fuel right and that is converted into heat and then to kinetic energy right. So if we are going to look at efficiencies we have to consider what is the input energy the input energy is the chemical energy that is stored in the fuel or propellants. Now a part of this not all of this chemical energy is going to be usefully converted only a part of it is going to be usefully converted to the power that we require or the thrust power that we call it at now you know from thermodynamics that you cannot have a system that can operate without with only one reservoir right you need two reservoirs one high-temperature reservoir and one low-temperature reservoir and heat must be lost to a low-temperature reservoir. So here also we have to lose heat to a low-temperature reservoir so because of the cycle the particular cycle that we follow we are going to lose certain amount of energy due to this let me call this as unavailable energy okay so this is because you have to lose heat to any cycle that you take even if you take Carnot cycle reversible Carnot cycle the efficiency cannot be one right because you need to lose heat to a low-temperature reservoir and therefore even if you look at the Carnot cycles you will have something like this Carnot cycle efficiency is ? is given by TH or 1-TL by TH right this is temperature of the low-temperature reservoir and this is the temperature of the high-temperature reservoir so it has to lose heat to a low-temperature reservoir and therefore this cannot be 0 therefore you have the efficiency is less than 1 so this component takes care of that now all the remaining energy that is available cannot also be converted to useful propulsive power part of it goes in the stream is lost in the stream as kinetic energy I am utilized okay and only the remaining portion is the useful in producing the propulsive effect now we will design define something known as thrust power we will define something known as thrust power okay thrust power is nothing but f x VA and this is the portion that we are talking about now what is the unutilized kinetic energy or the power lost in the jet that is given by mass flow rate m. a into this is the power lost in the jet so if we are looking at defining something known as propulsive efficiency this will be propulsive efficiency will be nothing but this divided by this entire quantity okay fine so that is FVA divided by FVA plus m. a Ve minve square – VA square by 2 so now how do we go about further simplifying this we know that thrust is F is equal to nothing but m. a into Ve – VA so we can use that here okay Ve – VA square – VA square we can use that as this is nothing but if I take out m. a it will be Ve plus VA by 2 so I will get propulsive efficiency if I take this as now I can cancel out all the F's so if I cancel out all the F's I will be left with propulsive efficiency is equal to 2 VA divided by sorry there was a small mistake and you did not correct me the power lost in the jet is not this Ve – VA whole square so if I take that into account I will get here if I take out this I will get this changes to Ve – VA by 2 fine okay so now I will get Ve plus VA I will define a ratio called R which is nothing but ratio of velocities given by VA by Ve I will define R which is ratio of velocities and using this I will get if I put it propulsive efficiency in terms of R I will get 2 R by 1 plus R and similarly I can write thrust also in terms of R as F is nothing but m. Ve into 1 – R now let us see how this varies with R what will when will this be maximum when will ?p be a maximum when R is 1 right and R is 1 it will be a maximum and if you look at this expression here thrust will go to 0 when R is equal to 1 so let us plot that this has to be m. A Ve so if I plot F by m. A Ve and ?p on the same axis y axis versus R how does ?p vary let us say this is R equal to 0 this is R equal to 1 and this is 1 here this is the variation of ?p what happens to F F is as R keeps on increasing F decreases so you get a straight line and it goes to 0 at this point okay so this is the variation of so if we go beyond R of 1 then the thrust produced will become negative and therefore has no meaning right has no meaning beyond this point and R less than 1 is the regime that we are looking at and if you look at this R is equal to 1 it goes to 0 so we cannot be at this point we have to be somewhere here such that you have us sufficiently large thrust and your propulsive efficiency is also high okay now if you remember when we were discussing about turbo props I made a mention that turbo props are devices wherein m. A is increased and the velocity differential is reduced in order to get the same thrust you can do it two ways you can either increase m. A or you can increase Ve-va if you increase Ve-va then the propulsive efficiency goes down okay so you are turbo props will have higher propulsive efficiency because of this so if you have R of around 0.44 you will get ?p of around something like 74% okay this is propulsive efficiency for aircraft engines let us look at what is the propulsive efficiency for rocket engines I do not think this is the right number for aircraft engines I will this is for rockets for aircrafts if f is equal to 70 kN and va is equal to 800 km per hour and Ve is equal to 1800 km per hour possible because the nozzle is choked at a higher temperature right so you have higher speeds at the exit then you get a propulsive efficiency of around 61.5% with this okay now let us look at what is the relationship for propulsive efficiency for rockets now again for rockets propulsive efficiency is thrust into velocity that is thrust power divided by thrust power plus kinetic energy lost in the jet okay now here we know that f is nothing but m. x Ve right so if we substitute this expression here we get ?p is equal to m. Ve mine Ve va if I bring the two from the denominator here divided by 2m. Ve va plus m. Ve- Ve2 plus Va2-2 Ve va okay now this and this cancels out right 2m. Ve va with a – sign and 2m. Ve va with a positive sign so these two will cancel out and you can also cancel m dots so you are left with Ve2 plus Va2 okay and if we use the same definition for Rs r is equal to Va by Ve we will get ?p is equal to 2r divided by 1 plus r2 and similarly if I were to write f in terms of r I will get f is equal to there is no r right it is independent of Va so therefore it will not change right so this is one of the key parameters here the thrust of a rocket motor is independent of the ambient velocity or the velocity of the vehicle okay so you get m. Ve or f by m. Ve is always 1 irrespective of whatever is the velocity Va that it is travelling at it still continues to produce thrust whereas in an aircraft engine or an air breathing engine you saw that if the velocity ratio was greater than 1 then the thrust produce would go to negative values or it would mean that there is a drag on the vehicle right here it is independent of it and it is always this quantity is always 1 okay. Now where will this be a maximum propulsive efficiency how do you determine that well very simple just differentiate with respect to r you will get 2r into 0 plus 2r minus 2 into 1 plus r2 whole square this is nothing but 4r square minus 2 plus minus 2r square so what should go to 0 for this to be a maxima the derivative must go to maxima or a minima the derivative must go to 0 so if you put this to 0 you will get the value of r that is r square is equal to 1 or r is equal to 1 is the value how do we know that this is a maxima you take a second derivative right that should be if it is a maximum what should be the second derivative should be negative if you take a second derivative this would be what will this be okay find that out now r is 1 and notice in the earlier case r equal to 1 we got thrust 0 for air breathing case now for rockets it will still be f by m dot ve would be still 1 so it is independent of the ambient velocities or the vehicle velocities so if I plot it in a different color chalk so propulsive efficiencies for rockets are typically slightly higher than that of air breathing engines because of the term 1 plus r square here if you remember it was 1 plus r for everything engines r being less than 1 it will always this term will be smaller compared to 1 plus r therefore you will find propulsive efficiencies for rockets will be slightly higher than those for air breathing engines so propulsive efficiency is higher why shouldn't we use rockets as the question we will try to look at okay now coming back to the question that I asked you right at the beginning suppose we were to use a rocket motor to run torpedo whether that would be useful or to get the exhaust from the rocket motor run a turbine and connect that turbine to the propeller and let the propeller do the work on the fluid with that be used okay let's see that now let's assume the mass flow rate m dot to be around 0.7 kg per second now if this were to make this were used as a gas generator and made to power a propeller the thrust produced would be typically something like would be around 16 kN at something like 80 km per hour remember water is you are looking at a torpedo and water is what kind of a fluid incompressible fluid right or compressibility effects are very very small you need to go to very high velocities in order to get to that right so we would be typically operating at very low velocities here so now this is the thrust that is produced if it is if this kind of flow rate is given to a propeller now let's say we were not doing that and we were giving we were directly throwing out the exhaust this is the situation wherein you have a gas generator or a combustor here and this is made to run a turbine turbine is connected to a propeller and the propeller powers the vehicle forward okay now instead of this you can also opt for another system wherein it is just like a rocket motor just only the combustor and you have a CD nozzle right let me call this situation one and I will call this situation to here if I have the same mass flow rate going through just a convergent divergent nozzle I will get a thrust this depends on exit velocities for a rocket motor if you remember it depends on exit velocities VE so F is nothing but m dot VE in this case VE is around 2300 meters per second so I get a thrust of 0.7 into that is somewhere around 1.6 kN okay if you look at this this number is 1 tenth of this right how did this happen the answer lies again in propulsive efficiency if you look at the R and propulsive efficiencies if you look at R and propulsive efficiencies for the two situations for this one the actual VA would be something like 22 meters per second okay so use take R as VA by VE that is 22 by 2300 around 0.01 and if you calculate propulsive efficiency you will get ?p is equal to 0.02 so most of the thrust or most of the useful power here is lost in the as the kinetic energy of the exhaust gases okay it is not used usefully so it will always be better if you do this propulsive efficiency wise so if you were to make a statement from based on this you will say that it would be ridiculous to use rockets for torpedoes right but Russians do have a top secret torpedo that they call as squall it is pronounced as squall which uses a rocket motor and therefore it goes at a speed of 100 meters per second or 360 kilometers per hour and its range is something like 6 to 4 kilometers very short range right and it goes very fast the other torpedoes comparable ones are something like have a very small velocities compared to this torpedoes speeds are of the order of less than half of this okay this is because the other torpedoes use the first kind of mode and a squall uses this kind of so it can go at a very high velocities but remember if you go at very high velocities the drag inside water will be enormous so they use something known as supercavitating flow wherein they produce gas bubbles in front of the vehicle if this is the vehicle if this is the vehicle and this is the exhaust they produce they have a gas bubble generator and produce gas bubbles all over this place right in front of it so that the drag on the vehicle is less because now you are looking at something like a two phase flow it is not pure water so the viscous drag will reduce and therefore they can make this kind of velocities right otherwise typically you do not find anybody using scenario to to power a torpedo because a lot of useful power is lost in only the kinetic energy of the jet okay now let us move on and look at what is the overall efficiency okay again going back to our figure this was unavailable energy you had unutilized kinetic energy and lastly you had the propulsive effect right in this was chemical energy stored in the QLR propellant so if I went to calculate thermal efficiency what should I look at what is thermal efficiency here propulsive efficiency was this quantity divided by this quantity so similarly what is thermal efficiency thermal efficiency is if you take output power as FVA that is this part plus this part and you divided by this one right this gives you the thermal efficiency remember I also earlier mentioned about Carnot cycle efficiencies and why this loss is there right so and the chemical energy stored in fearless or the input power is m.f x Q right this is this portion and we are looking at this portion right but in addition to this we also need to add one more term there is a power that is required to carry or transport the fuel remember these are flying systems and they cannot be powered like systems on ground they have to be lifted up and carried along with the vehicle so there is a power that needs to be added to this and that is m.f VA square by 2 this quantity is small if you look at aircraft engines compared to the compared to this this is small for aircraft engines but it is really not small in terms of rocket engines this quantity will be large for rocket engines okay so from this we can find that thermal efficiency thermal efficiency is FVA plus m.a right this is the expression for thermal efficiency right and this expression holds for both aircraft and for air breathing as well as non-air breathing we will stop here in the next class we will look at what is the overall efficiency and what role does it play in terms of the range of the aircraft and other things okay we will stop here thank you.