How can we express the thrust provided by a rocket engine? In rocket engine we have propellants that are injected at high pressure in the combustion chamber. They burn, any excess gas are expanded in a nozzle. They are expanded from P_c, pressure chamber, until a pressure at the outlet of the nozzle P_e. The mass flow rate and that is ejected and the gas here will have a velocity, V_e. But first, we have the first term, which is the product between this mass flow rate and the velocity, V_e. This is what we call the momentum thrust. Moreover, we have this pressure P_e, which in general will be different than the pressure of the atmosphere, P_a. In fact, this pressure P_e only depends on P_c and on the design of this nozzle. In fact, this nozzle is a convergent divergent nozzle. We will have Mach equal 1 in the throat. From basic term of gas dynamic relation, we can demonstrate that the ratio between P_c and P_e only depend on what we call the expansion ratio. Where epsilon equal the ratio between the outlet section area on the nozzle and the throat section area. Whereas P_a is the pressure of the atmosphere. Therefore this gas here that arrives at the pressure of P_e at the outlet of the nozzle will see a pressure P_a and this will induce another term in the thrust, which is the pressure thrust equal P_e minus P_a times A_e, where A_e is the area of the outlet section on the nozzle. This is the pressure thrust. Now, P_a is not constant. We have a launcher that will follow a trajectory. The pressure of the atmosphere will change in time. That means that also our thrust will change in time because of this pressure thrust. That can become very important. It can represent up to 30 percent of the total thrust. When P_e equal P_a and the pressure thrust is 0. Well, we say that the nozzle is adapted. It can be shown that for a specific nozzle, this is the best situation with respect to performance. But of course, we can just have one altitude where P_e equal P_a. In general, it will not be the case. Let's introduce now the equivalent velocity. Well, the equivalent velocity is the velocity that we should have at the outlet of an adapted nozzle that provide the same thrust that we have here. That means that we have this equality, that trusts equal M dot times the equivalent velocity. Or the equivalent velocity is just the ratio between thrust and mass flow rate. The specific impulse now. First, let's introduce the total impulse. The total impulse is the integral between zero and T, time of thrust times d t. Now, the specific impulse is the ratio between the total impulse and the weight of the propellant used in order to generate the thrust. That means that it is ratio 0 t, T dt divided by g0, 0 t m dot d t, where g0 is the gravitational acceleration. We can have two possibilities. Either we have T and m dot constant, which is not very frequent, or we can consider that we have small amount of time in which we can consider t and m dot constant and take them out from the integral. We arrive to this more common expression, which is ISP equal T over g0 times m dot. We see that there is a very straightforward relation between the equivalent velocity and the specific impulse. That is, the equivalent velocity is the specific impulse time g0. Lets add a couple of details about the equivalent velocity. To do so, it is very useful to introduce another expression. It is the mass flow rate. As we mentioned, the nozzle in a rocket engine is chocked. That means that the Mach is one in this throat. We can express the mass flow rate in a chocked nozzle, it is equal gamma time Pc At divided by the square root of R Tc divided by W. Pc, pressure chamber, At area of the throat, Gamma is just a complicated function of the small gamma, which is the ratio between the specific heats. It's something around 0.66, 0.65 depends on the burned gas. Then we have Tc is the temperature in the chamber, R is the universal gas constant, and W is the molecular mass of the gas that will depend on the ratio O over F. Well, of course, let's remind that all these expression are simplified. 1D expression that are derived from gas dynamic relation. That means that either we are considering 1D flow or we are considering average quantities in each section. Anyway, taking this relation of the mass flow rate, replacing it in the expression of the equivalent velocity or Isp, this is what we get. The equivalent velocity equal the thrust divided Pc At times square root R Tc divided by W, and we have our gamma. We see here that we have two terms. The first one is called the thrust coefficient. By convention, we indicate it work, Cf. It is the ratio between the thrust and the product Pc At. It can be interpreted as the thrust provided by the nozzle. That means that Pc At can be used as a phrase we would have without a nozzle, just with the pressure here and the float area. Indeed this coefficient depends on the ratio gamma of the specific hits. It is very related with the design of the nozzle and not on what happens inside the combustion chamber. It varies between one and two, roughly. The other term is called the characteristic velocity, and by convention, it is indicated with C star. It has a unit of a velocity, like the equivalent velocity, and C star only depends on what happened inside the combustion chamber. Here we see that we have the temperature of the combustion gas and we have the molecular mass of the burnt gas that appears and Gamma, which is a function of the Cp over Cv. C star heavily depends on the choice of propellants. Actually, as the highest Isp does as a consequence and we usually try to maximize it and by doing so, we try to maximize actually the ratio between the combustion temperature and the molecular mass of the burnt gas. Actually what we can see here is that this equivalent velocity has two terms. One, Cf, which is associated with the efficiency of the nozzle, and one, C star which is associated with the efficiency of the combustion. But of course, these are ideal expression. In reality, we will have efficiencies because we have losses, heat losses, non-complete combustion so we have efficiency in terms of C star, or because we have some divergence losses in the Cf or whatever. In reality, we will have expression that are more Veq equal one efficiency of Cf times Cf one efficiency of C star times C star. Now that we have seen the different expression of the thrust and for ISP, let's analyze them. What is the thrust that we need? Well, we need a thrust high enough for our launcher in order to take off and this is a requirement. If we are limited in terms of the equivalent velocity, we can always increase this thrust, we just need to increase the mass flow rate. This is what actually strap-on booster are needed for. If you look at Ariane 5, you have solid strap-on booster. They are there in order to provide thrust because they eject a lot of mass flow rate. But their Isp is not very high. In order to increase the Isp, it's much more difficult, we need to increase either the Cf or the C star. Both things can be accomplished by increasing the pressure chamber. Actually increasing the pressure chamber will slightly increase the C star because of the combustion process that can be more complete and increasing the pressure chamber can allow to increase the Cf because we can either have higher expansion ratio or lower throat. In either case, we can increase the Cf and increase the Isp. Now that you have all those expression under our eyes, we can examine them a little bit. First of all, the thrust: why we need the thrust? We need a thrust in order to takeoff, for instance, launcher. That means that in that case, we need a thrust which will be higher than the weight of the launcher. Here we have an equivalent velocity because we have made the several tries. A way in order to increase this thrust would be to increase the mass flow rate. We can put more engine or we can have strap-on booster that can help to take off, to increase the thrust. In order to increase the Isp is another story. In fact, the Isp or the equivalent velocity tell us how much propellants we need to burn in order to have a specific thrust. That means that it's an efficiency of our engine. We cannot just increase the mass flow rate to increase the Isp, we need to do other things. In particular, looking at this expression of the equivalent velocity, we can have some clue. We can increase the equivalent velocity by increasing either the C star or the Cf. Increasing the pressure chamber, we can increase both. In fact, if we increase the pressure chamber, we can increase Tc, and if we increase the pressure chamber, and in particular, we increase the ratio Pc over Pe, we can increase the thrust coefficient.