 So in the previous lecture, we had discussed how to estimate the burnout velocity of a rocket vehicle as well as the burnout height using a phenomenological approach where we balance the forces and we consider the thrust as a force being supplied as a reaction force. And then in the previous lecture, we started a different approach of the launch vehicle dynamic estimation where we consider the exhaust as the factor producing the thrust. So, we will continue from the same discussion. So, let me first draw the rocket that we were discussing last time. So, this is the rocket that we were discussing. This rocket is flying with the velocity u and this is the positive x direction, this is positive y direction. The forces acting on this rocket one is the drag force acting in the negative x direction against the flight path opposite to the flight path. Then the weight of the vehicle acting on this c g is given by m g. The lift is acting normal to the flight path given by l and we had considered that at time t mass of the vehicle was equal to m. The velocity of the vehicle was equal to u. Then what we have said is that the rocket is flying at a particular condition. We look back at this rocket after time t plus delta t that is after a small time interval delta t. What are the changes that has happened in this rocket? That is what we want to see. So, we say that during this small time interval there is exhaust which comes out of this rocket. Let us assume that the mass of that exhaust which is coming out during this small time delta t was equal to d m and the velocity of this exhaust is equal to u e. So, then the mass at time t plus delta t at time t plus delta t, the instantaneous mass is equal to the initial mass minus the mass that has gone out and the velocity of the vehicle has increased by a small amount delta u. Therefore, the new velocity is u plus delta u and we have considered the exhaust velocity to be equal to u e and let us say exhaust mass, exhausted mass rather let me write it exhausted mass is equal to d m. So, this is the problem we have been discussing in the previous class and we have shown that the change in the momentum of this vehicle total momentum of this vehicle along the x direction during this small interval of time d t is equal to m d u minus d m u e. This is what we had derived in the previous class where m is the mass of the vehicle at time t d u is the increment in the vehicle velocity during this small time d t d m is the mass exhausted from the vehicle during this small time interval and u e is the exhaust velocity. So, this is the expression that we had derived in the last class. Now, let us continue with the discussion coming back to this diagram and now let us just revisit the different forces that we had talked about. Previously we had identified four forces where thrust was a reaction force appearing separately and then we considered only the vehicle not the exhaust. Therefore, the rate of change of momentum was just m d v d t. Now, we are considering the exhaust also therefore, we do not consider thrust as a separate force. So, then the forces acting on the vehicle are only this three lift drag and accelerate the weight. So, the weight of the vehicle is due to the gravity. So, this is equal to force because of gravity. We have or is the same expression will continue where m is the instantaneous mass g e is the acceleration due to gravity at the c level r e is the radius of earth and h is the height of the vehicle at particular instant t. Now, this two together is the local acceleration due to gravity. So, we can write this as m times g. So, that is the gravitational force or the weight. Then the drag again we have seen this before the drag is equal to half rho infinity u infinity square s c d or s we can also write as a f where c d is the drag coefficient a f is the frontal area or weighted area u v is the free stream velocity of the incoming air rho infinity is the density of air at that altitude or density of the medium may not be here. What we can see from here is that if we are outside the atmosphere then rho infinity is 0. Therefore, the drag is 0 as we go up high up in the altitude density reduces. Therefore, the drag reduces. So, initially at the launch pad we have substantial drag, but as we go up the drag continuously keeps on reducing. So, this is the expression for drag the lift is given by half rho infinity u infinity square s c d. So, once again here c l is the lift coefficient a f is the frontal area u infinity is the free stream velocity rho infinity is the density. This lift and drag are essentially the aerodynamic parameter depends on the aerodynamic design as well as the operating altitude and operating speed. So, these are nothing to do with the rocket designer per se this is something the aerodynamic should provide to the rocket designer. So, these are the two values that we will be considering. However, one more thing that will come into picture in this case which did not appear in the previous analysis which is the forces exerted by the exhaust pressure the pressure forces. Because there will be a pressure differential may be a pressure differential here. So, we have discussed that there are various type of expansion possible. We can have an under expanded nozzle, we can have a ideally expanded nozzle, we can have an over expanded nozzle. We have also discussed in the previous lectures that the thrust produces maximum for ideal expansion because at that condition P e is equal to P a that is the exit pressure is equal to the atmospheric pressure. So, therefore, there is no contribution from the pressure forces and we have also shown that for either the over expansion or for the under expansion the thrust is going to be less than the ideal expansion. We have already discussed those things we have also discussed that typically what we would like to have is over expansion not under expansion because of the fact that when the vehicle is launched from the ground the pressure is low. As it goes up in the altitude the pressure continuously decreases and as the pressure is decreasing if we start with different type of expansion there is a monotonic decrease in the expansion. So, but if we start with a point where let us say P e is less than P a this is under sorry when we start with under expansion P e is less than P a is under expansion. We start with this condition P e is less than P a as it goes up P a falls the atmospheric pressure falls and then at one point of time it will become equal to P a beyond that P e will be greater than P a. So, we have a thrust variation something like this whereas, if we start with this condition and then this P a is falling. So, we have a continuous drop in thrust. So, therefore, this is a preferred condition P e less than P a then at certain instance we will reach this condition where P e equal to P a. So, this is a digression from what we have been discussing. Now, let us this is over expansion expansion is more yeah this is over expansion condition. So, that is why we usually consider over expansion, but ideal case will be this where we get the maximum thrust. So, now since we are considering the exhaust separately there should be a contribution from this pressure force if the expansion is not ideal. So, that also now needs to be accounted for in the momentum equation. So, the fourth force in this case is equal to the pressure forces and that is equal to P e minus P a times a e this will be applicable if it is not ideally expanded. If it is ideally expanded P e will be equal to P a. Now, let us look at what are this forces doing the drag force is along the along x direction or is along the direction of flight we consider x as the direction of flight. The gravitation force. So, first let us look at the forces acting along the x direction. One is the drag force second is the a component of F g a component of gravitational force will also be acting along the x direction and that component will be equal to let me say F g x which is equal to M g cos theta according to this diagram where this angle is theta. Then the net and this pressure force is also acting in the x direction which is equal to P e minus P a times a e. Now, the net force then along x direction is equal to this pressure force P e minus P a times a which will be acting in this direction upward in the positive x direction minus the drag force. So, this is minus d and this term the gravitational component is also acting in the minus x direction. So, that will also be minus. So, this is M g cos theta. So, therefore, this is the net total force external force acting on the rocket vehicle. Now, from Newton's second law of motion this is equal to the rate of change of momentum. So, now, let us use Newton's second law of motion. So, then first of all total sigma F along x direction is equal to the total change in momentum in the x direction is equal to this. So, this can be written as M d u minus d M u e. So, this is the application of Newton's second law of motion on this rocket vehicle. Now, let us simplify it little bit. What we can first what we will do is we will divide both sides by d t. So, sorry just one more thing this is the net force acting which is equal to rate of change of momentum. This is the change in momentum the rate of change is momentum will be d t divided by d t. So, therefore, let me rewrite it again d M d u minus d M by u e d t. This is the rate of change of momentum is equal to the sum of external forces. This is the governing equation that we have. Now, let us look at this equation little more closely. So, first of all let us look at this term d M change in mass d M is nothing but equal to M dot d t where M dot is rate of change of mass. So, this is M dot d t where d t is the time interval. Now, this is equal to the initial mass was M. So, this is d M d t times d t and the mass of the vehicle is decreasing. So, there will be a minus sign here. So, d M is minus d M d t d t. So, this is the expression for d M where 1 second like to reiterate that the change is occurring because of the propellant flow. So, this is nothing but propellant flow d M is the propellant flow that brings about this change in mass. So, then what we can do is if I look at this expression we will have M d u. So, now going back to this equation M d u is equal to P e minus P a times A e plus this term d M d t d M by d t times u e is M dot u a. It had a negative sign when we take it to the right hand side this becomes positive and then after that we. So, this term here will be equal to d M d t is equal to M dot d t. So, this term here d M u u by d t is equal to M dot u a. It has the negative sign when we take it to right hand side it becomes positive. So, this is equal to M dot u e minus track minus the gravitational term and all this times d t. So, this d t I am taking to the right hand side. So, this is the expression for M d u. We can further simplify it and write an expression for d u all alone. The change in velocity is equal to first of all let me look at this two term together. What is this? We know that M dot u equivalent equal to M dot u e plus P e minus P a times A e. Therefore, these two term together is M dot u a. So, we write it here as M dot u e M dot u equivalent minus d M d by M d t. Well, d is the drag M is the instantaneous mass minus g cos theta d t. That is the acceleration due to gravity. Now, let us look at M dot term. M dot is this will have divided by M, because we are taking this M to the right hand side. Now, M dot by M from this expression we write M dot by M equal to d M by M. If I divide both of this by M. So, M dot there will be a d t here first of all, because this d t is also coming in here. So, M dot d t by M is equal to minus d M by M. So, now, we take this and put it back into this equation. This simplifies to minus u equivalent d M by M minus d by M d t minus g cos theta d t. So, now, we have got an expression for d u. As you can see that this includes the equivalent velocity. It is not in terms of the thrust directly. It is in terms of the equivalent velocity and the mass change is already been incorporated in this equation. So, now, we work with this equation and now let us go back to the same similar cases that we had discussed for the other approach in the previous lecture and try to get the expression for velocity increment as well as the height that can be achieved. So, let me first rewrite this equation. d u is equal to now minus u equivalent d M by M minus capital D by M d t minus g cos theta d t. This is the general expression. I like to point out here one thing that this is the expression in the positive x direction, velocity increment in the positive x direction. We have lift is not appearing here because lift is in the y direction. So, we do not consider the lift at present. So, this is the generic equation which can be applicable to any rocket vehicle. Now, let us do the case studies. Case one, first let us consider that the drag is 0 and we also consider that the gravitational force is 0 or g is 0, acceleration due to gravity is 0. So, for this case then from this equation this term goes to 0, this term goes to 0. So, what we are left with a simple equation d u equal to minus u equivalent d M by M. As you can see this expression is much simpler than what we have done before. And now we can just integrate it from time t equal to 0 to some time t and we get the velocity change. So, when we integrate this between time t equal to 0 to some time t then we get this is equal to delta u minus u equivalent d M by M integration is just log. So, this will be log M by M naught where M naught is the initial mass of the vehicle at time t equal to 0. And M like in our discussion here is instantaneous mass of the vehicle at time t. So, the velocity change therefore, this is now the expression that we have obtained. So, starting from initial we can now get expression for any time, this negative sign is quite odd. So, what we will do is we will just flip over and we can write it as u equivalent L n M naught by M. So, as we can see this is a much more simplified expression that the difference between this expression and what we had obtained earlier is the fact that there we had considered the thrust acting at an angle little vector thrust. Here we are not considering that we are saying that the exhaust is coming in the x direction negative x direction that is why those complexities do not arrive here. So, but we have discussed that why do we need to have that small variation that because of the stability. So, now let us continue with this discussion and see that what we get after this. So, let us consider a practical total burnout at time t equal to 0 we had initial mass M naught. And then the rocket is launched till the complete burnout we want to find out how much velocity change is occurring. So, at time t equal to t b which is the burnout time the mass is equal to M f which have been writing as M b bus, but here since we are talking about a different approach. So, just to differentiate it I will write it as M f is the final mass. So, the velocity increment during this entire operation then is equal to u equivalent L n M naught by M f. Now, if you recall in one of the previous classes I have defined this mass ratio and that mass ratio is given as M r which we have defined. So, let me write it here M r is equal to M naught by M f which we have defined and sorry it is M f by M naught. And we had also defined the propellant mass fraction zeta which is equal to 1 minus M r this we had defined before. So, now coming back to this expression then we get delta u is equal to u equivalent L n M naught. L n 1 upon M r according to this definition we can also write it as equal to minus u equivalent L n M r we can write it like this. Now, notice one fact here that M r is always less than 1 because fuel mass is always less than the total vehicular mass. So, M r is always less than 1 therefore, this term is going to be negative. So, this becomes positive delta u is positive and since zeta equal to 1 minus M r. So, therefore, M r is equal to 1 minus zeta. So, this can also be written as minus u equivalent L n 1 minus zeta. So, now we have got the single stage rocket dynamics the velocity change in terms of the mass ratio or the propellant mass fraction. I would like to point out here that typically for any rocket these are the factors that are given to the designer. So, you have to work with these factors only that is why it is important to express everything in terms of these ratios and particularly becomes important when you go to multistage because therefore, different stages will have different numbers for these ratios and you have to work with all of them together. So, therefore, it is important to work with these ratios the mass ratio or propellant mass fraction or structural coefficient which will come later everything. So, this is the first case now we have discussed when the drag is 0 no gravity then this is the exit velocity expression which is expressed in terms of the mass ratio as well as the propellant mass fraction. Now, continuing from this then let us consider second case so, this is our basic governing equation. So, I will just retain it here and look at a second scenario in this case we will consider the gravity we will still neglect the drag and consider the gravity. So, case 2 case 2 is then f g is not equal to 0 d is still equal to 0 we still consider no drag, but now we are considering the acceleration due to gravity. In that case coming back to this equation then this term goes to 0 this term remains now this term as we can see is g cos theta d t when we integrate this equation this term gives me this and integrating this between time t equal to 0 to d t to t gives us g cos theta t that is it nothing else changes. So, we can bring it here and finally, we get delta u is equal to u equivalent minus u equivalent l n m r. That is this term after integration is giving me this minus g cos theta t and if you are burning up to t b up to the burning time t b. So, this gives me g cos theta d t b one point like to mention here that during your flight g may vary cos theta may vary. So, if this two are varying then we have to account for that also, but when we are integrating this then we can take the average value since we are integrating. So, we can take the average value of g cos theta. So, average g cos theta times t b gives us then the expression of velocity increment with an incremental velocity with the with the effect of gravity also. So, del u is the change in velocity. So, here t b is our burning time which we have discussed before here t b is our burning time and cos theta particularly this cos theta is the integrate at average value of cos theta that is that the angle that the flight path mesh with the vertical is the average value of that. So, this is actually not over the entire thing but only the cos theta average cos theta. Now, this integrate this averaging is valid if we are talking about short thrust intervals. If we are talking about law interval of course then we cannot consider the average value then what we have to do it at every instant we have to consider the attitude and then integrate it. Now, gravitational acceleration h so far as long as we are close to the gravitational close to the earth surface it does not vary much. But once we start moving away from the earth then it starts to vary with h the attitude because we have shown that g equal to g equal to this we have already talked about this. So, as we go away from the earth surface h becomes important then g will start to drop. So, therefore this value of g as long as we are close to the earth surface it will be a constant. But beyond a certain height cannot consider this to be a constant then in this integration we consider the variation of g as well and then accordingly we integrate. So, therefore that can be considered. Now, one point I would like to emphasize here we have mentioned that this term is always going to be positive because m r is less than 1. So, this term is always positive g is always positive it cannot be negative right acceleration due to gravity is always positive theta is bound between 0 and 1. So, that is also positive time is already increasing. So, this everything here is positive. So, therefore this term is also positive. So, this is positive minus a positive quantity. Therefore, the effect of gravity is to actually decrease the acceleration due to the increase in velocity. So, we have to ensure that this term should be large enough to compensate for this loss because of gravity. So, therefore in the design state that is why we are writing all in terms of this parameters that are very equivalent for the mass ratio should be chosen in such a way that we get a substantial velocity increment that is the m finally to get a high velocity which is not much affected by the variation in gravity. So, that needs to be. So, therefore the thrust must be high enough to compensate for the gravity effects here this is the thrust the term appearing here is from the thrust. So, this is the second case that we have talked about let us now look at the another scenario a third case. Now, in the third case what we will consider is that now we do not consider the drag is absent let us say it is the flight is inside the atmosphere. So, we consider the drag as well. So, in case 3 in case 3 what we will consider is that the drag is also present. So, we have the gravitational effect we have the drag as well. So, case 3 is considering the drag also. So, we have defined drag as half rho u square f c d this is how we have defined the drag. Now, let us see what is happening here density is changes with altitude because we consider the atmosphere to be the perfect gas consider perfect gas. Then we know that for perfect gas equation of state pressure equal to rho r t. Therefore, rho which is the density is equal to pressure by r t. Now, temperature as we go up in the atmosphere initially decreases remains constant then increases again then remains constant decreases like that at different levels of the atmosphere temperature varies. The pressure monotonically varies from the earth surface up to the edge of the atmosphere the pressure continuously decreases. The net effect is the density continuously decreases as we go up high in the altitude. So, that there is a change in the altitude the density also changes. Therefore, considering everything same the drag will be reducing as we go up and up in the altitude. Second point here is that u as we have just discussed del u is always positive. Therefore, as del u is positive u is also changing and increasing. So, u changes with altitude. Now, I like to point out here one thing I said that the del u is positive, but that is only during the powered flight. If we cut off the engine this term goes to 0. Then this term will take over and reduce del u and it will come to 0 velocity and that will come back again when we talk about the maximum height. So, during the powered flight del u is positive. Therefore, u also changes with altitude. So, once again what we are saying is the density is decreasing with altitude, but velocity is increasing. Considering the aerodynamic design is same c d is constant let us say area is constant. Then the drag will be increasing because the dependence is u square and the variation in velocity is much higher than the variation in density. So, therefore, the drag continuously changes with altitude. So, d changes with altitude. Now, here altitude represents essentially the passage of time because the variation in altitude is because we are increasing increases with time. So, therefore, in other words we can say that it changes with time. So, now if we come back to this expression then this drag here is a function of time. Now, we have to integrate this equation. So, d equal to d t. So, coming back to this equation this dependence is not a linear dependence unfortunately as you can see that everything is coupled. So, this is our governing equation now where minus d t by m d t minus g cos theta d t. This is what we have to integrate now. So, this is what is not very easy to integrate because we do not have a well defined closed form term for d as a function of t. We do not know how drag is changing with time. It is not very easy to figure out. So, typically what is done is a round about approach in solving this. So, the common practice is this. Let us assume that everything remains constant. Let us assume that everything remains constant during a small increment of time d t. So, we assume that this theta remains constant of d t. Gravitational acceleration remains constant in this small interval. The d remains constant. The mass remains constant. Everything is constant. The equivalent is also constant even in this time. Then now we can very easily integrate this. We can get a value of this entire right hand side at a small time. Then slowly we keep on changing it. So, numerical integration that is what we can do. So, what we can do is then we calculate this value of drag calculate d as a constant because during this small interval our density is constant, u is constant. So, we can calculate this value of d during this small interval of time. Once we have done that, now we put this value back into this equation. Now this is appearing as a constant. So, now after that we integrate this and get a velocity increment d u. So, next we calculate d u with constant d. We increment this d u with constant d. We calculate this value of d u. Now what we do is in the next time step, we replace this u with u plus d u and the corresponding Q and D altitude also we estimate. Replace the density with that and recalculate it for the next time step. So, again we put that value here and integrate this. That is why that is how we keep on integrating in time or marching in time till we get the final value. And also one more point here is that when we are doing this, we also keep on changing this theta whatever change in theta is required that we keep on doing because that comes from the component balance. So, we keep on upgrading this term and do a time marching and estimate the final value. That is how we estimate the final velocity increment when we consider the drag and gravity as well. So, this gives us the most complex scenario which can also be done but has to be done numerically. I like to point out here one more thing that in all this analysis we have not talked about y direction. If we look at y direction then unless we provide a thrust vectoring there will be a non-zero lift. So, there will be a tendency of the vehicle to move in this direction. So, in order to get zero lift we have to give a vector a small vector to the thrust which we have discussed in the previous lectures. So, therefore in order to provide in order to have zero lift the thrust must be given at an angle to the u direction or the x direction as we have already discussed. This balance the normal component of gravitational force otherwise the gravitational force will not be balanced and it will start to wear off its intended path. So, now let us see if there is no lift and we do not just give vector then there is a unbalanced force acting like this and that will take it downward. So, it will come down. So, therefore in order to compensate for this gravitational pull we have to give a slight vectoring to the thrust. So, we put thrust at a particular angle here and now coming to the other way then we can actually achieve various type of flight stable flight by providing thrust vectoring. If we do not provide thrust vectoring there will be a tendency to go like this. If we provide thrust vectoring it will go like this let us say after a point we want to bring it down we we have a different thrust vectoring it comes like this. So, a cruise missile where we want to change the orientation or attitude of the flight path we can do that by thrust vectoring very easily. So, therefore thrust vectoring is an important method in the flight of rocket vehicles. So, with this we come to an end of our discussion of this method. Now, what we will do is in the next lecture we will go back to the single stage rockets and look at different performance parameters like the maximum height etcetera. So, here we have estimated the velocity change. Next we will look at the burnout time and burnout height etcetera for single stage rocket which we are discussing so far. Thank you.