 So, welcome to this lecture on rocket propulsion. Before we proceed, let us first recap what we have discussed so far. We have been discussing the vehicle dynamics for the last few lectures. So far, we have been focusing on single stage rockets and we have defined some parameters. One of such parameter is the mass ratio, which is given by the final mass by the initial mass. Then we have defined various type of masses, that is ML is the payload mass, then MP is the propellant mass, MS is the structural mass, which the structural mass includes everything except this two, except the payload mass and propellant mass everything else is included in the structural mass. Then the total mass at the beginning is the sum of all this and at the burnout all the propellant is supposed to have been used up. Therefore, what remains is only the structure and payload. So, the final mass is nothing but the sum of structural mass and payload mass given by MS plus ML. We have seen this last time, then we had defined another parameter R, which is inverse of mass ratio. Therefore, R is equal to M naught by M f. Now, if we combine these two, we get R equal to M naught amount ML plus MS, we had done it last time. After that last time we defined some more parameters, one of the parameter was payload ratio, which is designated by lambda and payload ratio is defined as the ratio of the payload mass divided by all the initial mass except that of the payload. So, it is ML upon M naught minus ML. Now, again coming back to this expression, if we subtract the payload mass from the initial mass, what remains is the structural mass and the propellant mass. So, therefore, lambda is equal to ML upon MS plus MP, we defined lambda in the last class. Similarly, we defined another parameter, which is structural coefficient designated by epsilon and we had defined epsilon as again the structural mass divided by all other mass except the payload, we defined it like this. Therefore, this again comes out to be equal to MS upon MS plus MP, we have done, we have defined these two parameters in the last class. After that, we had combined these three definitions R lambda and epsilon and we had shown that R is equal to 1 plus lambda upon lambda plus epsilon, this is what we showed in the last lecture. So, we have done it in the last lecture. Now, let us proceed from here. First of all, let us look back at the structural mass. Structural mass as we have said is everything except the propellant mass and the payload mass. So, now, if we look at a practical system, a rocket essentially is the payload and the engine and tanks, fuel tanks. So, therefore, if we take out the propellant and the payload, what remains actually is the engine and tank. So, therefore, the structural mass is mass of the engine plus mass of the tank where the subscript E stands for engine and subscript T stands for the tank. So, therefore, the structural mass is just the sum of the engine mass and the tank mass. Then the payload mass now can be written in terms of this. What we can see from this expression is that the payload mass is equal to initial mass minus the structural mass minus the propellant mass. So, payload mass is equal to M naught minus M s minus M p and we can put M s equal to M e plus M t like we have done here. So, therefore, the payload mass is equal to M naught M e plus M t minus M p. This gives us an expression for the payload mass. Now, what we are interested in is finding out the payload ratio, not the payload ratio, the payload fraction that is this is called payload fraction M L by M naught. So, this is equal to then first of all what we do is we take this expression for M L and then we divide it by the initial mass M naught. Then after this division what we will see is that this term is equal to 1 minus we write M e by M naught as the engine mass ratio fraction. Similarly, we have the tank mass fraction. This thing let us multiply and divide by the propellant mass and then the propellant mass fraction upon M naught. So, the payload mass fraction we can write like this we have just expanded this. Now, let us look at the expression for R which we have defined here. R is nothing but M dot by M f and M f is equal to the initial mass minus the propellant mass according to our definition. We can write it like this. Hold this equation for the time being. Let us go back now to the expression we had derived for the velocity increment. We had what we have done is we have derived expression for various cases. We have considered first of all no lift for most of the cases and they have we have considered no gravity, no drag, etcetera, etcetera. For the time being let us look at the simplest case. What was the simplest case? When we had no drag and no gravity. Now, if I look at the velocity increment for this then we have proved that this is equal to delta u, u equivalent L and R. This we have proved in the previous classes. This equivalent velocity R is this ratio as defined here and u equivalent is delta u is the velocity increment. Now, from this then what we can do is if we integrate this we can get an expression for R in terms of delta u and u equivalent. So, that will be equal to e to the power delta u by u equivalent not u e. So, integrating this we can get the expression for R as a function of velocity increment and u equivalent velocity. So, now let us put it back here as equal to R equal to delta u by u equivalent. Then now what we can do is first let us invert this. So, we can write m p up sorry m naught minus m p divided by m naught is equal to e to the power minus delta u by u equivalent. The left hand side of this expression can be written as 1 minus m p upon m naught. This is the left hand side of this expression is equal to then e to the power minus delta u by u equivalent. So, we are just simplifying this expression. Now, what by doing this what we can do is with what we can get is we can get an expression for m p by m naught that is the propellant mass fraction which is a very important design parameter. So, what we have shown here there is that 1 minus m p by m naught is equal to e to the power minus delta u by u equivalent. Therefore, we can write m p by m naught which is the propellant mass fraction is equal to 1 minus e to the power minus delta u by u equivalent. Now, what is the significance of this expression? This expression is what the designer will be using. Let us look at this equation m naught is the initial mass. First we have to lift this initial mass and we have to attend certain velocity at the end of the burnout. So, that delta u is dictated by the vehicle mission dynamics. So, the mission director will specify how much delta u will be required at the end of burning of the propellant. Then you have already chosen a propellant. So, you have chosen a propellant u equivalent is fixed because that is a function of I S B specific impulse. So, once you have chosen the u equivalent delta u is specified by the mission requirement then from this equation first of all we know the propellant mass fraction. And since m naught is the initial mass then we know how much propellant we need to carry for achieving this mission. So, this is a very important equation that tells us how much propellant needs to be carried by the vehicle to achieve a particular delta u and a particular fuel which will be dictating the u equivalent. Now, first once we have got this let us come back to this expression for the payload mass fraction. So, we had shown it in that board that the payload mass fraction is given as 1 minus m e by m naught minus. Now, this term here we will simplify little more. So, let me put it as m t by m p and m p by m naught and then minus m p by m naught. Now, this m p by m naught we have derived here. So, between these two if I take m p by m naught common then this equation can be written as 1 minus m e by m naught minus 1 plus m p by m p times 1 minus e to the power minus u by u equivalent. We get this expression. Now, this expression then tells us that see the engine mass again is something that is fixed tank mass is something that is fixed. Now, this propellant mass we are estimating from here delta u and u equivalent is now known. So, we cannot tell that for the given propellant that we are carrying and the initial mass how much payload we can carry. So, once again the mission goal is to carry this payload and give it that increment in velocity delta u. From this we can estimate are we able to carry that payload or not or in other words what we can do is if m l is given m naught is given and these are given we can find out how much propellant is required or if we need to change the propellant is the u equivalent good enough to give us that velocity increment or do we need to improve on the design of this or make the tanks lighter. Various things essentially can come up all the the mission and design requirement will come up by looking at this equation. So, therefore, this equations are very very critical. Now, I have said at the beginning that the tank and tank mass and the engine mass are the function of the are the part of the structural mass they are included in the structural mass. So, now if I use that and define our structural coefficient again epsilon then by that definition m t plus m e which is equal to the structural mass m s divided by the propellant mass plus the structural mass will come in here. So, this is the structural coefficient defined in terms of the tank and engine mass as well. So, now let us look at this expression what we will do is we will take this equation and then put it back along with this and this we will combine all three equations. Once we combine all three equations we will get expression for the structural coefficient in terms of other masses and the mission requirement. Now, this is our mission requirement this ratio. So, this is equal to m t pi m p 1 minus e to the power minus delta u by u equivalent plus m e by m naught as you can see here this plus this essentially comes from this term here that is m e plus m t by m p. So, this is essentially nothing but this divided by m naught. So, we are essentially what we are doing is we are dividing and both the numerator and denominator by m naught that is what we are doing and then m t by m naught can be written as m t by m p m p by m naught like we have done here right and m p by m naught we have already expanded in terms of this. So, this is what we are doing. So, if I write the expression here continue with this expression this is equal to then 1 plus m t upon m p plus m e by m naught. Now, I have given you two homework so far this is the third homework derive this expression. So, now coming back to this equation or this expression what do we see? We see that the structural coefficient depends on what are the parameters. If we consider a particular mission the tank mass and the engine mass are fixed because we have chosen that then the structural coefficient essentially will be dictated by this term initial mass is also dictated fixed. So, therefore, what we see is that the structural coefficient is a function of delta u and u equivalent. So, structural coefficient depends on how much velocity increment we want and what is the equivalent exhaust velocity. On the other hand we have shown that the equivalent velocity is a function of I s p that we have shown before that equivalent velocity is a function of I s p specific impulse and specific impulse depends on thrust per unit mass fuel flow rate. So, it is a function of the fuel that we are considering or the propellant that we are choosing. So, therefore, this u equivalent equivalent velocity then is a function of delta u and propellants. So, the structural coefficient also depends on the mission requirement and the propellant chosen to achieve this mission. So, this expression gives us that functional dependence. Let us now look at a practical case and see particularly how the structural coefficient depends on these two parameters. For that what I will do is I will plot the variation of structural coefficient as a function of delta u for two different fuels or two different propellant particularly we will consider fuel as the one of the major propellant is the fuel. So, let me plot structural coefficient along the y axis and the velocity increment along the x axis. Let us consider two fuels one is a hydrogen fuel. If I plot the variation of delta u for hydrogen fuel it is like this it varies like this this for the hydrogen fuel. If I plot the same for a hydrocarbon fuel there is a steep drop and then it is like this this is a hydrocarbon fuel. So, this is what has been seen experimentally that the structural coefficient variation with respect to the delta u that is the velocity variation or in other words if you vary delta u how structural coefficient is going to vary for different fuels that is what we have shown here. Now, let us see that what we see from this plot. We see that for a given delta u you look at anywhere any location here. For given delta u we see that the structural coefficient is greater for hydrocarbon as the fuel compared to sorry for hydrogen as the fuel compared to hydrocarbon as the fuel. So, for any given velocity increment the required structural coefficient is going to be greater for hydrogen than hydrocarbon fuel. Now, hydrogen as we know is much lighter than hydrocarbon. So, essentially what is this graph is showing is that for a given velocity increment the structural coefficient for a lighter fuel is greater than that for a heavier fuel. So, this graph shows that the structural coefficient for a lighter fuel is going to be higher than that of the heavier fuel. Now, once again what is the significance of lighter fuel and heavier fuel? If we consider a lighter fuel as mass flow rate is less. If we consider two fuels which have similar say heating value one is lighter other is heavier then the specific impulse defined as thrust per flow rate. Now, as long as you are adding the same amount of energy the thrust is going to be same if you have the same rocket design. So, if I look at the specific impulse specific impulse is defined as this is what the definition of specific impulse is. So, what we see is that specific impulse is inversely proportional to the fuel flow rate. So, therefore, a lighter fuel will be giving us higher specific impulse because for the amount of same amount of same energy it is supplying it is lighter or the weight is less. So, less amount is consumed less mass is consumed. So, therefore, we know that the specific impulse for a lighter fuel is greater than the specific impulse for a heavy fuel. So, then if I look at these variations if the specific impulse is greater what happens to our equivalent velocity everything remaining same equivalent velocity is higher. So, therefore, we can also conclude that the equivalent velocity provided by the lighter fuel is going to be greater than that provided by a heavier fuel. So, since specific impulse is higher the equivalent velocity for the lighter fuel is greater than equivalent velocity for the heavy fuel. Therefore, now what we have already discussed that the structural coefficient is function of both the delta u as well as equivalent. So, for the same delta u we see that the lighter fuel has higher equivalent velocity. Therefore, the structural coefficient is higher for lighter fuel compared to a heavier fuel. So, that is the physical explanation of this behavior. So, this is the first point. Let us look at second point here what we see if I just take any of these graphs any one of them any particular fuel we are choosing what we see is that at lower value of delta u when the velocity increment is less then there is a decrease in structural coefficient as delta u is increased. But beyond a certain amount value of delta u the structural coefficient becomes constant. So, let me write it down for lower delta u we see that as delta u increases the structural coefficient decreases and for higher delta u higher velocity increment structural coefficient is independent of delta u then it is a function solely of u equivalent as we have just discussed its function solely of u equivalent. So, coming back to this why is this first let us look at the higher delta u how do we achieve a higher value of delta u by burning more propellant for a longer period. So, higher delta u means higher propellant mass m p since we are talking about the initial mass being fixed for all the cases. So, the initial mass is fixed also we do not want to change the payload. Then where do we accommodate the higher propellant only way is by cutting out the structural mass. So, therefore, at higher m p means there is a reduction in structural mass. So, higher delta u can be achieved for keeping everything same by reducing the structural mass. And therefore, what we are seeing is that primarily the structural mass is much less than the propellant mass. Hence, since we are achieving this structural coefficient variation rather the delta u variation by varying m p only and the structural coefficient is typically inversely proportional to the propellant mass. Therefore, at very high values of delta u epsilon is almost constant because the marginal dependence decreases. We have already reached such a high value we cannot change it further structural coefficient further. So, the structural coefficient is almost constant. And once again coming back to this discussion that why for the lighter fuel we have higher structural coefficient. What we see here is epsilon is typically inversely proportional to the propellant mass. Because if you have more propellant of course, the structural has to be satisfied. So, structural coefficient is less. So, again coming back to this then if we take a lighter fuel propellant mass is less. Therefore, the structural coefficient we can go to higher value. So, we can work with higher value of structural coefficient if the propellant mass is less. And that is why we see this increase. So, therefore, what we can conclude from this discussion is that first of all if we are operating with a lighter fuel like hydrogen. We can get satisfactory performance that is higher delta u at a higher structural coefficient that is what this discussion show. However, if we are working with heavier fuel then the structural coefficient is less in order to achieve the same delta u. Now, what is the significance of that as far as the rocket designer is concerned? If you are allowed to have higher structural coefficient you can make the structure stronger. So, your factor of safety improves. So, therefore, the reliability of the mission or the robustness of the rocket improves if you are allowed to work with higher structural coefficient. Because finally, the structure is the one that will have to withstand all the other loads external loads acting on it. So, therefore, working with a lighter fuel is helpful for the designer to design because they are allowed to design a stronger or heavier structure and that helps in improving the reliability of the rocket. Second point here is that beyond a certain value of velocity increment we cannot we have to work with a fixed structural coefficient. Structural coefficient does not change. However, we can increase the value of delta u. However, initially as we increase the delta u structural coefficient sharply decreases. So, the structure has to be made lighter and lighter in order to attain the higher velocity. But once we are here then it is independent of structural coefficient. Now, this point here that for a wide range of delta u typically the structural coefficient does not change is something that we have to keep in mind. Because later on when we go to multi stage optimization this is something that we will be using that typically the structural coefficient becomes independent of delta u or any other parameter when we go to higher delta u values. So, this discussion shows us the dependence of structural coefficient on delta u. Now, let us go back to the discussion that we had regarding the flight dynamics for various cases. What we are trying to find out is how much velocity increment that we can get. So, we are providing the vehicle with certain amount of thrust and the vehicle is flying and it is getting certain velocity increment after certain burn out time. It is carrying certain amount of propellant which is getting burned and it is getting the velocity increment. What we have done during the discussion so far is we have considered different cases and made some assumptions in those cases. Now, let us look at how this assumption affect the final performance. So, what will be the effect of those assumptions in the final design. So, for that let me draw the variation of delta v versus the thrust. So, I will draw two curves one is here. In the x axis we will have thrust divided by or non-dimensionalized by the initial weight. So, thrust non-dimensionalized by the initial weight then we take any rocket with any initial configuration it will be applicable. On the y axis we will take velocity increment divided by isp g naught. So, this is the variation we are looking at. Let us for the sake of convenience consider a particular mass ratio m naught by m f. Let us say this mass ratio is 8 and let us also consider payload fraction. This is payload fraction right because if I say m naught minus m f no. So, this is the propellant fraction. So, m naught minus m f gives us the propellant fraction which is equal to 8.875. So, these are the conditions for which I am taking a single stage rocket and I am plotting the variation of delta v essentially with respect to the thrust that I am providing. So, first of all let us consider the free space. For the free space this is g equal to 0 case. Say this is 1 2 3 etcetera this is 0.5 1 1.5 2 etcetera. So, what we are seeing is that when we consider this particular rocket once we have these values fixed everything fixed then the variation of delta u with thrust actually will be a constant line right. Why? Because delta u is just u equivalent L and R. R is fixed here u equivalent is this raised to this therefore, it will not depend on thrust. So, in the free space no matter what amount of thrust we are providing it is a function of only u equivalent and the mass ratios therefore, it will be a constant delta delta v it does not depend on thrust. So, this is a very important observation in the free space outer space and that is why in outer space we do not need huge chemical rockets. What we need is something with higher ISP right. So, that is that will give us the required variation because of the fact here because in the outer space we do not have acceleration due to gravity we do not have drag. So, this is the case we are considering. Then the second case let us say is for this is case 2. So, case 1 we have said here is free space where G is equal to 0. Case 2 we consider horizontal flight when G is equal to G naught we are considering horizontal flight. For horizontal flight initially as we increase the thrust there is an increase in delta v, but gradually it reaches the free space condition asymptotically as we increase the thrust a lot because what happens is the vehicle is flying horizontally the weight is acting vertically downward right. If you keep on increasing the thrust this essentially becomes insignificant compared to the thrust, but initially when the thrust is less then it will have a tendency to bring it down or the delta u which is we are interested in this direction will come up come down, but as we keep on increasing the thrust becomes much much more than the weight then it will not have that impact. So, therefore, it asymptotically reaches this value. Now, the third case we consider is vertical flight including now this is the vertical flight which is G equal to G naught again. If I consider vertical flight it will go like this when we are talking about vertical flight G is always acting downward always try to slow down. We are increasing the thrust that will try to compensate for the effect of gravity, but gravity is always present. Therefore, it will never reach this condition and it will always be less than this, but as we again keep on increasing the thrust the effect of gravity becomes less and less dominating that is why we get this variation. And the same thing if I look at for different let us say ratios like instead of this if I look at another case where m dot by m f is 5 and m dot minus m f by m dot is 0.8. So, the same plots if I plot then let me plot it on the same graph, but with a different color. We will see that the free flight condition will be less because now your r is less therefore, the free flight velocity is going to be less. At the same time every other parameter now will be less. So, this is 1, this will be 2 and 3 is even less somewhere here. So, what we see is that as r is decrease and the propellant fraction is decrease there is a overall decrease in the velocity increment which is essentially according to the equation that we have derived that as the thrust is the velocity increment will increase as r increases. So, therefore, this is in agreement with what we have derived theoretically. So, therefore, again since as I said delta v is something that is the mission requirement which will become apparent when we talk about space dynamics. Therefore, in order to achieve that we need to choose a particular value of r and once that is fixed we need to choose a particular value of structural coefficient that I have just discussed. So, if the mission is given if the propellant is chosen every other parameters will come up from this analysis. So, then that is what the rocket design is that what should be the mass distribution of various components and then we actually go into build it, but the first step is to find out the mass distribution to achieve that mission. So, this brings us to the end of our discussion on single strain rocket performance, but before we stop what I would like to do is I would like to solve a problem I would like to solve a problem that will essentially focus on the single stage performance that we have discussed so far. So, let me define the problem let us consider a single stage rocket this is a single stage rocket. Let us say that the this is our payload let us consider that the payload mass is 100 kg. Let us consider that the ISP for this rocket is 450 seconds. Let us consider that the structural coefficient or the structural factor I will define what is structural factor is 0.2 and ideal delta V is equal to 2000 meter per second. Now, first of all what is ideal delta V comes becomes evident from this plot. Ideal delta V is the maximum velocity that we can achieve and it should not depend on thrust. So, it corresponds to which case G equal to 0. So, ideal case corresponds to no drag no gravity the outer space. So, for this case you are asked to find out calculate the lift of mass which is m dot this is equal to what that is the question. So, this is essentially what we have been discussing so far. So, let us now look at what is given is ML equal to 100 kg ISP equal to 450 seconds. Structural factor let us say we call it S which is equal to 0.2 is defined as m s upon m s plus m p. So, this is equal to m s upon m naught minus m l we have to get delta we are given that delta V equal to 2000 meter per second. So, first of all we know that for this case delta V is equal to u equivalent l n R right this we have proved and u equivalent is equal to ISP times G E. So, ISP times G E l n R is equal to m dot minus m f. So, first of all let us look at this expression little more closely. Let us say we define a parameter l which is the propellant mass fraction l equal to m dot by m l. So, from here we can write that just a second m dot by m l equal to 1 by l right and m f equal to m dot minus m p right and our s is sorry let me do it from here s equal to m s upon m l m dot minus m l right. So, therefore, let me first look at this one let me do 1 minus s. So, 1 minus s is equal to 1 upon m s m dot minus m l right. So, this is equal to m dot minus m l minus m s divided by m dot minus m l right. What is this m dot minus m l minus m s m p? So, this is equal to m p upon m dot minus m l. Now, let us multiply this by l. So, let me multiply this by l. Then what I get is this is equal to m m p upon m dot minus m l into l is defined as m l upon m naught. Now, to this let me add s. So, let me add s to this plus s. So, this will be equal to m p upon m naught minus m l into m l upon m naught plus m s upon m naught minus m l right. What we see is that this is common in this two. So, after I simplify this, this will be equal to delta v we have got like this. So, delta v will be equal to i s p g e l n 1 upon l 1 minus s plus s where l is our propellant mass fraction and s is the structural factor. Now, from here then we get l 1 minus s plus s is equal to exponential minus delta v upon i s p g e. Once we simplify all this, you will get the overall mass is 184 k g. But let me try it out little differently also. Let me try it out little differently. Since I am starting to do with this, this is the final solution, but I will do it little differently. Once again let us look at this ratio m dot upon m f. So, m dot upon m f is equal to m dot upon finally, what remains is the payload mass and the structural mass. And we have defined epsilon as this m s upon m s plus m p. So, this is our epsilon s is our epsilon. So, this is equal to m s upon m dot minus m l. Therefore, we get m s equal to epsilon times m dot minus m l. You can get this expression. So, m s plus m l is equal to epsilon times m dot minus epsilon times m l times m l. So, we are adding m l to this equation. Then this can be written as epsilon m naught plus 1 minus epsilon times m l. So, let us take this now and put it back into this equation. So, this will be then m dot upon epsilon m dot plus 1 minus epsilon m l. Now, let us divide both of them by m dot. So, we get this is equal to 1 and m l by m naught is equal to l. Therefore, m dot by m f is equal to 1 upon epsilon plus 1 minus epsilon times l. So, now let us take this come back to this equation and put it back here. So, what we get is this is equal to instead of doing it here, we can also start from here right away our m l is given epsilon is given. So, I can write this as equal to 0.2 m naught plus 0.8 m l. So, now, if I take this and put it back into this equation what I get is if I put this back into this expression for delta v, I get 2000 which is our ideal delta v equal to 450 which is the is p times 9.8 acceleration due to gravity equal to multiplied by l n m naught 0.2 m naught plus 80 right. Now, in this equation my only unknown is m naught, I can easily solve for this. So, after solving for this I get m naught is equal to 184 kg. So, as we can see that we get the expression that we have been looking for and then we get the initial mass that will give us this velocity increment which will put a 100 kg payload at two certain velocity increment. So, with this we complete our discussion on solid single stage rockets. Next class what we will do is we will start with multi stage rockets, we will define various parameters for multi stage rocket. We will see what is the advantage of multi stage rocket to begin with why we need multi stage rocket and then what are the various parameters and then how to optimize the performance of a multi stage rocket. After that then we will go into the space dynamics that is the why do we need certain velocity increment delta v. So, that then we will talk about the mission requirements. So, thank you very much we will stop here today.