 Good morning. So, let us continue our course on rocket and spacecraft propulsion. For the last couple of lectures, we have been discussing the performance of ideal rockets. We have listed the assumptions that are required. Then we have defined certain parameters. One of those parameters for the thrust coefficient defined as f t upon a star p c naught, where f t is the thrust produced by the rocket, a star is the throat area for the converging diverging nozzle. Here, we are considering the rocket to be like this, a combustion chamber and a converging diverging nozzle. So, this area is a star the minimum area throat area. The combustion chamber conditions are p c naught and t c naught that is the stagnation pressure and stagnation temperature in the combustion chamber. So, we have defined the thrust coefficient like this. Then we have proved that the critical mass flow rate or the choked mass flow rate and since the nozzle is going to be choked. So, this is the mass flow rate is equal to a star p c naught square root of gamma upon square root of r t c naught 2 upon gamma plus 1 to the power gamma plus 1 upon 2 gamma minus 1. So, this is the mass flow rate. As we can see here, the mass flow rate is the function of the throat area and the chamber conditions. Now, in order to increase the mass flow rate, since the throat area is constant, only way to increase the mass flow rate is by changing the chamber conditions. If you reduce the combustion chamber temperature, the mass flow rate is going to increase or if we increase the combustion chamber pressure, the mass flow rate is going to increase. So, the possibility for increasing the mass flow rate and we know that the mass flow rate will dictate the thrust, total thrust because thrust is m dot u e plus the pressure term or m dot u equivalent. So, in order to increase the thrust, one way is to increase the mass flow rate. Now, thus we have also proved in the last class that the thrust coefficient can be made independent of every other parameter, but the pressure ratio and gamma. So, we have proved this in the last class, gamma to the power gamma minus 1, 1 minus p e upon p c naught to the power gamma minus 1 by gamma to the power half. This is the momentum thrust and the pressure thrust given as here p e is the exit pressure at the exit of the nozzle, p a is the ambient pressure, a e is the exit area that is the area here. And therefore, we see that the thrust coefficient is function of gamma, the pressure ratio and the area ratio. However, we have shown before that area ratio is a function of Mach number, therefore it is a function of pressure for the isentropic pressure ratio. Therefore, the area ratio is function of pressure ratio. So, we can eliminate this area ratio from here and we can get it completely as a function of pressure ratio and we had discussed that. So, what we have shown that C f is a function of the pressure ratio and gamma. Now, once the design is fixed, p e is fixed then p e by p c naught is a function of area ratio. So, that becomes a fixed ratio then in order to change this we have to change the pressure, but we have all. So, what we have shown is that as we increase the pressure the thrust coefficient is going to increase, but we have also shown that it cannot increase to infinity, there is a maximum limit. When this ratio tends to 0 that is the exit pressure is very small compared to the stagnation pressure we get the ultimate thrust coefficient. After this we had defined the characteristic velocity as p c naught m star upon m dot. I have said in the previous class that the thrust coefficient essentially dictates the performance of the nozzle and the characteristic velocity dictates the performance of the combustion chamber. So, with this then we have proved that the total thrust is the product of mass flow rate, thrust coefficient and characteristic velocity. And based on this we have also proved that the specific impulse will be product of C m and C star. We have proved that in the last class. After that we had derived an expression for C star and we have shown that C star is equal to this. So, we have shown that in the last lecture and we had stopped here at the end of the last lecture. Now, we have the two expressions for specific thrust sorry the thrust coefficient and characteristic velocity. We see that the I S P is the product of this two. From our mission requirement from the very beginning we have been saying that we want to maximize the I S P. So, that our equivalent velocity will be higher. Now, the two parameters that dictate if I look at this expression C star there is primarily a function of T c naught and gamma. So, gamma is present in both and of course the molecular weight. So, for a given fuel it is basically fixed. So, now again gamma for a given fuel or a given composition will be fixed because the final composition of product will dictate what will be the value of gamma. So, the operating conditions that we have are essentially T c naught and P e by P c naught. Once again I have said that for the given area ratio P e is fixed for a given P c naught. So, therefore this can be replaced by P c naught. So, the two parameters that we have are the stagnation chamber pressure and stagnation chamber temperature. Now, coming to this expression or rather this expression first of all we can increase thrust by in three ways looking from this equation. Either we increase the mass flow rate or we increase the C f or we increase the C star. Let us say we want to increase the mass flow rate and the parameters that we have are P c naught and T c naught. So, here is our expression for mass flow rate. We see that in order to increase mass flow rate we have to either increase the stagnation pressure or decrease the stagnation temperature. If we increase the stagnation pressure C f is increasing. So, in this equation then both this will increase this will increase and C star is not dependent on stagnation pressure to that extent. So, therefore the thrust is going to increase if we are going to increase the stagnation pressure. Other way of increasing the mass flow rate is by reducing T c naught. As we reduce T c naught as we can see from this expression C f is not that much affected, but if you reduce T c naught our C star is going to reduce. So, characteristic velocity is going to reduce and the net effect can be that the thrust may reduce. So, that also we do not want. Therefore, the better way of achieving a higher thrust is by increasing both P c naught and T c naught. However, there is a limit to P c naught first of all the structure puts a limit because that will be the force exerted on the wall of the rocket by the gases. So, structures put a limit to that and secondly the temperature if it is very high there is a possibility of thermal defects on the wall you need lot of cooling etcetera. And secondly very high temperature will lead to dissociation of the gases and when the gases start to dissociate gamma is going to change which is something we do not want and dissociation actually is an endothermic reaction. So, it takes away some energy. So, therefore, the overall energy is going to reduce because of that again the thrust produce will be less. I will discuss this dissociation effect again. So, bottom line is although we can increase both P c naught and T c naught to get higher thrust, but there is a limit to it that is why we cannot keep on increasing the thrust as much as we want. Now, this discussion where we have defined C F and C star now will allow us to attempt design of rockets. So, that is what we are going to talk next simple rocket chamber thrust chamber design and nozzle design and then I will solve a problem today on that. Let us now look at simple rocket rocket thrust chamber. The rocket thrust chamber is the rocket motor and the nozzle because that was produce the thrust. Let us consider that the inlet of the combustion chamber is 1, the exit of the combustion chamber and inlet of the nozzle is 2, the throat is given by star and the exit of the nozzle is E. These are the different stations. Let us also consider that the amount of flow going in is m dot in. This is the combination of both fuel and oxidizer all the propellants. Let us also consider that because of chemical reaction the heat liberated is q r. So, that is the q that is added to this flow rate and then this is the same flow that goes out. So, first of all m dot in is equal to m dot out because the mass flow must be conserved. If I draw the T s diagram for this process T s temperature entropy diagram of this process at the beginning of the thrust chamber at 1 we are here. Then what is happening here in the combustion chamber? In the combustion chamber we consider the process are to be isobaric because it was considered constant pressure process that is what was proved by for isentropic process. So, we consider the process in the heat addition process of heat addition to be a constant pressure process or isobaric process. So, therefore, in the combustion chamber between 1 to 2 there is a constant pressure heat addition. So, if I plot it plot it it will be going to 2 and the pressure remains constant at p c naught. Now, because of the heat addition there is an increase in temperature. So, the temperature here at the at 2 is now p c naught. So, this is the chamber temperature. Now, this rise in temperature depends on the amount of heat that is added. So, this will be equal to q r upon c p that can be proved from energy balance. Now, after the combustion chamber in the nozzle there is an expansion. We are considering the expansion to be ideal. So, is isentropic. So, it is goes down vertically. So, this is our point e exit. This is the pressure at the exit given by p e you are not commenting about p e, but is isentropic. So, it is the ideal expansion let us say. In between somewhere it crosses the critical condition also p star. So, from 2 to p star is isentropic p star 2 e is isentropic. So, this is the cycle that the flow goes through from 1 to 2 to star 2 e. So, once again 1 to 2 is a constant pressure heat addition and 2 to e is isentropic expansion. These are the two processes. Now, let us see what is happening. First of all the energy is added in the combustion chamber. So, let us first do an energy balance for the combustion chamber. So, energy balance between 1 and 2. The total heat added is m dot which is the mass flow rate times q r. This is the total heat that is added and then this is the flowing process. So, from thermodynamics you know for the flowing process from first law of thermodynamics since there is no work the only work is the p d v work. Therefore, the heat added is equal to the changes enthalpy stagnation enthalpy. So, the stagnation enthalpy change now the stagnation enthalpy change is nothing but the stagnation enthalpy at 2 minus stagnation enthalpy at 1 that is the exit of the combustion minus the inlet of the combustion. So, this is equal to m dot h 0 2 minus h 0 1. Now, we have assumed the walking fluid to a perfect gas for ideal rocket which means it is thermo-interfect, calorically perfect. So, therefore, h is equal to c p t and we consider c p to be constant. So, therefore, this can be written as m dot c p t 0 2 minus t 0 1 and that is what we see here. That is what I have written here q r by c p is equal to t 0 2 minus t 0 1. So, from this equation now I can derive the value estimate the value of t 0 2. So, if I do that from this equation I will get t 0 2 which is the stagnation temperature at the exit of the combustor which will be our t 0 c t c 0 is equal to t 0 1 plus q r upon c p. So, if the fuel and oxidizers are given the propellants are specified then for a particular propellant combination there is a particular heating value q r. So, that is a known quantity then from the product c p is also a known quantity. So, if these two are given initial temperature is given then we can estimate the final temperature. Now, in the nozzle the flow is isentropic and we have proved that for an isentropic flow for isentropic flow H naught is constant. Therefore, H 0 2 is equal to H 0 E stagnation enthalpy remains constant. So, stagnation enthalpy at the inlet of the nozzle is equal to that at the exit of the nozzle and H 0 E is equal to the static enthalpy at the exit plus the kinetic energy term u e square by 2. Therefore, we get u e square by 2 equal to c p t 0 2 minus t e. So, now in this equation and c p is known t 0 2 is known if we can estimate t e we can estimate the exit velocity u e. We are considering the nozzle to be isentropic. So, therefore, the process for the nozzle flow will be considered to be isentropic. So, we can use the isentropic relationship and then for isentropic expansion in the nozzle we get the exit velocity is equal to square root of 2 c p t c naught 1 minus p e upon p c naught to the power gamma minus 1 by gamma. Here t c naught is equal to t 0 2 as I have said in the stagnation chamber combustion chamber stagnation temperature. Here what we have done is we have written this as c p t 0 2 1 minus t e upon t 0 2 and then t 0 2 we have put equal to t c naught t e by t naught 2 is replaced by p e by p c naught to the power gamma minus 1 by gamma. This is the isentropic relationship. So, since we are considering the expansion to be isentropic we can write it like this. So, for the isentropic condition nozzle this is the expression for the exit velocity. So, this is first time now we have an expression for the exit velocity u e. This we have also seen in Gaster by the same expression. So, this will now give us the exit velocity at the exit of the nozzle. What we will do is we will simplify this little more. We will express it in terms of the known quantities which are given to us. So, in order to do that what I will do is t c naught I will replace by this equation. So, if I do that I will get the exit velocity is equal to square root of 2 c p t c naught 2 equal to t naught 1 plus q r upon c p 1 minus p e upon p c naught to the power gamma minus 1 by gamma. Now, we are doing an engineering estimate. So, as engineers we always make some approximations. Now, let us make some approximation. Let us look at this term t 0 1 plus q r upon c p t 0 1 is the incoming flow temperature. Typically, for Gaster mines when the flow enters the combustor it has been it has gone through the compressor. So, is fairly hot about 400 500 Kelvin temperature will be there, but for the rocket it is coming directly from the supply usually it is not hot. So, temperature is low. So, t 0 1 is typically low. So, and then combustion occurs there is a rise in temperature. Temperature can increase by about 2000 up to 2000-3000 Kelvin quite a bit of temperature rise. So, this delta c p which is a temperature rise is much larger than the incoming temperature. So, essentially we can write t 0 1 is much smaller than q r upon c p. Now, if we make this assumption then from this term we can neglect t 0 1 then u will be simplified to u e will be simplified to as a function of it is approximation to q r. Now, once we make that assumption you have a c p here and a c p here. So, this two will cancel off. So, it will be approximately two q r and only this term. So, 1 minus p upon p c naught to the power gamma minus 1 by gamma. Now, once again we are noticing something we have eliminated the temperature again. What the exit velocity is a function of is the heating value. So, it is a propellant property. It is a heating value of the propellant and p upon p c naught, but there is a cache to it. Here we are considering that the entire available chemical energy is being converted to the thermal energy or heat. So, we are assuming that the entire q r is released that is why we are eliminating t naught because then t naught becomes a function t c naught becomes function of the total heating value. In reality it may not happen there is something called combustion efficiency all the energy may not be released. In that case the temperature comes into picture or what we do is we say the 90 percent energy is released. So, we multiply it with the combustion efficiency and then accordingly we can adjust to the requirement. Now, this is our expression for the exit velocity. Here we can also express the heating value as heating value per unit mole. If you do that then molecular weight will also come into picture like we had done earlier when we bought in the molecular weight in characteristic velocity. So, same thing we can do here also. So, if we consider q r per mole that is q r bar then my exit velocity will be equal to square root of 2 bar by molecular weight 1 minus p e upon p c naught to the power gamma minus 1 by gamma. Here this q r needs to be calculated using thermo chemical analysis which is part of the combustion analysis as can be done which we will do later. In the later part when we go to the combustion we will study how to estimate the value of q r. So, at present this is our exit velocity. So, now we have the exit velocity estimated. So, for a given operation now we can decide whether this whether a rocket is going to fulfill the mission or not. Now, to continue from here what I will do now is next solve a problem to make this concepts more clearer. Essentially what I am going to do is I am going to design a rocket now. So, when you are talking about design your boss will tell you that you have to produce certain amount of thrust. That is what the rocket designer needs then how do you design it that is the question. So, let me solve a problem now. So, the question is that we are supposed to design a rocket which means the thrust chamber as well as the rocket thrust chamber. So, rocket thrust chamber means the combustor and nozzle. So, we have to design a rocket thrust chamber. I will give you the conditions or the parameters under which we have to design. So, design of a simple rocket thrust chamber. Let us say that the propellants first of all is a very important parameter because this m bar comes here q bar comes here q r bar these are all propellant properties gamma is also propellant property. So, first let me say that the propellants are given it is nitric acid plus aniline a very commonly used both of them are very commonly used liquid propellants n rockets. So, nitric acid and aniline are the propellants. So, the chemical composition is this is these are the propellants that are used. Now, it is also said that the mixture ratio that is the ratio of nitric acid and aniline is equal to 2.75. At present this information we will not need because that will go into the chemical composition and chemical analysis. Here now we are not doing the thermo chemical analysis. So, we will not this need this information, but for the completeness of the problem we are providing that. What we are going to need is the combustion chamber pressure P c naught. So, stagnation pressure in the combustion chamber is specified to be 300 PSI A. I will give this problem in British units FPS units because of the fact that this is essentially an American problem. Typically this aniline nitric acid propellant are used by is used by Americans. So, American work with this system of units. So, I will give this system of unit. You can convert this also to SI units. So, P c naught is equal to 300 PSI A atmospheric pressure. Let us say this is the initial stage the first stage. So, therefore, the atmospheric pressure is the sea level atmospheric pressure which in absolute term is 14.7 PSI A. So, that is atmospheric pressure. Let us say that this rocket this is the most important parameter needs to produce 1000 pound thrust that is the thrust this rocket needs to produce. We will give some more parameters one is the constraint that the combustion chamber or the thrust chamber should have a cylindrical combustion chamber. So, the combustion chamber is cylindrical. The chemical analysis of this as I have said that it will come from the chemical analysis. So, from the chemical analysis of the product we get certain parameters for this mixture ratio for this propellant. So, the chemical analysis the chemical results give that the P c naught combustion chamber stagnation temperature is 4930 Fahrenheit which is also equal to 5390 Rankine. Then the exhaust molecular weight molecular weight of the exhaust or the burn product is given to be 25 pound per mole. Now, when you are designing this rocket you have to produce certain ISP you are designing it for that. So, the ISP is specified let us say the ISP is equal to 218 second that is what you need and again from the chemical analysis we get gamma equal to 1.22. So, for this now this conditions we have to design a thrust chamber that is the combustion chamber and the nozzle we will first start with the nozzle. So, first let us look at the nozzle configuration for the nozzle what is the important parameter it is the thrust coefficient. So, first for this case let us estimate the thrust coefficient we know that thrust coefficient is given as this plus this is the general expression for thrust coefficient. Now, here is a design problem we do not know the area ratio. So, first thing is we start with an assumption that is an ideal expansion. So, let us assume ideal expansion now with this assumption then P e is equal to P a therefore, the pressure term in the thrust coefficient goes to 0. So, we are left only with the momentum term and exit pressure is equal to the ambient pressure. So, this P e is equal to P a now in this equation the value of gamma is given P e is equal to P a which is given P c naught is given. So, I can directly calculate the thrust coefficient. So, if I do that so therefore, P e is equal to P a this implies the thrust coefficient after putting all the values will be equal to 1.41. Now, what does thrust coefficient give me why first we are estimating the thrust coefficient because we are supposed to design the nozzle. So, design essentially means we have to estimate the areas how do I estimate the area I get this thrust coefficient now let us look at the definition of thrust coefficient. Thrust coefficient is defined as the thrust by a star P c naught that is how I have defined the thrust coefficient. Now, let us look at this equation the thrust coefficient I have estimated already the thrust is given the stagnation pressure is given. So, only are known in this equation is a star the thrust area. So, I can solve for the thrust area from here as this now the values are given. So, I can directly solve for this this comes out to be equal to 2.36 square inch. So, 2.36 square inch is a my thrust area therefore, I can get d star which is the thrust diameter equal to 4 a star by pi this is equal to 1.73 inches. So, that is my thrust area for this rocket 1.73 inches. Now, once we have this and let us take the next step. So, first way now we have designed the thrust next let us look at the exit area. So, let me draw the schematic here what we have to design is this a star we have to design is this we have to get this exit area a e we have to get the length of this combustor l and the diameter of this combustor d that is what we need to design. So, first of all from here I get a star next step is to get the exit area a e in order to get the exit area we need to know the exit velocity. So, next let us calculate the exit velocity how do we calculate the exit velocity let us again look at what has been given to us what is given to us is the value of I s p specific impulse. So, now we use the definition of equivalent velocity which is equal to specific impulse terms times the acceleration due to gravity at sea level. Now, the equivalent velocity is u e plus p e minus p a upon a e by m dot but in this case we are considering ideal expansion. So, p e is equal to p a therefore, this term is 0. So, the exit velocity is equal to nothing, but I s p times g e I s p is given to me to 18 second g e is the acceleration due to gravity which is a known constant. So, I can get u e this terms out to be equal to 7 0 2 0 feet per second. Once I have this value next we need to calculate the exit area for the exit area what I will do is I will use continuity equation. So, from continuity equation exit density times exit area times exit velocity is m dot total mass flow rate. Therefore, the exit area is m dot upon rho e u e and what is m dot from the definition of thrust for this case from the definition of thrust f equal to m dot u e. So, I can replace m dot as f upon u e. So, in this equation let me replace m dot by f upon u e. So, I get f upon rho e u e square f value the thrust is given the exit velocity we have estimated. Now, we need to know the density how do we get the density for that what I will do is I will multiply and divide by rho e u e rho naught. Rho naught is density in the combustion chamber or stagnation density. So, this is something 1 by rho naught since we are considering the gases to be perfect gas rho equal to p by r t right. So, for perfect gas rho equal to p by r t. So, rho naught equal to p c naught upon r t c naught. So, I can put it here and rho e by rho naught by rho e can be given in terms of pressure ratio because is isentropic process. So, if I do that this finally comes out to be equal to f upon u square this is the value of r universal gas constant divided by the molecular weight r was coming there p c naught by p c naught p c naught by p a to the power 1 upon gamma. Here this ratio comes from the isentropic relationship relating rho naught by rho to p naught by p. Now, in this equation my thrust is known exit velocity have estimated this is the universal gas constant which is the known value this is the molecular weight which is given this is given this is given this is given this is given. So, everything here is given I can just plug in the values and get the final area this comes out to be equal to 8.6 square inch. So, now I have this area also I have this area I have this area also I have this area I have this area I can estimate the exit diameter the exit diameter d e is equal to square root of 4 a e upon pi. So, this is 3.64 inches about 4 less than 4 inches this is a small rocket we are designing. Now, we have this two parameters so now we have the nozzle configuration completely one thing the nozzle is essentially I am getting that so it throat area and the exit area I would like to point out here that the shape of the nozzle still we are not talking about because we have not discussed that later on we will come to the shape of the nozzle at present we are considering this is a shape nozzle, but we are not specifying what should be the shape. So, the nozzle part is done now let us look at the combustion chamber next. So, the combustion chamber what we need to know the velocity velocity in combustion chamber typically the velocity in the combustion chamber given by u c is much less than that in the nozzle which is quite obvious because combustion chamber conditions are almost stagnant whereas nozzle has a high velocity flow. Now, this is something that we assume is a design parameter that we assume. So, let us typical values of u c is between about 200 to 400 feet per second that is standard values of u c u c lies between 200 to 400 feet per second the combustion chamber velocity. For this problem let us consider or choose that u c is 250 feet per second reasonable value 250 feet per second velocity. Now, with this velocity then first of all we use the continuity equation for the first combustion chamber this is equal to rho c a c u c which is equal to m dot. Once again m dot can be replaced by u by u equivalent f by u equivalent as we have done here. So, if I come to this equation now I can replace m dot by f by u equivalent this u c will be equal to u v equal to u c this is rho c. So, everything here can be replaced I can get an expression for the area of the combustion chamber this you can write it as see this term f by u equivalent is my m dot this term r by m t c naught sorry this will be p c naught by p c naught this is my rho naught that is the density in the combustion chamber and this is u c right. So, we have m dot by rho c u c everything here is known. So, I can get the combustion chamber area this is equal to 20.5 square inch therefore, the combustion chamber diameter is about 5 inches that is the combustion chamber diameter. So, once we have estimated the combustion chamber diameter next we need the combustion chamber length. Now, length is something that depends on characteristic time and all and the combustion instability parameters. At present what I will do is we will consider a length factor which is typically given by the ratio of the combustion chamber volume to the critical area. So, let us assume combustion chamber volume to a star this typically a given parameter which has been evolved through experiences. So, typical value of this is about 60. So, let us assume this value is 60 a star we have estimated. So, this gives us the combustion chamber volume to be 141.6 inch cube. So, roughly about 140 inch cube therefore, the length of the combustion chamber will be volume of the combustion chamber divided by the area of the combustion chamber this is about 6.8 inches. So, if I draw the schematic now with the proper dimensions first let me write down this dimensions I will try to make it as much to the scale as possible. So, first let me write it that the length of the combustor I am getting is 6.8 inches the diameter of the combustor is about 5 inches the throat area sorry the throat diameter is was about 2. something right the throat diameter was equal to about 2.36 inches. No 1.73 inches and my exit diameter was 3.3 inches these are the 4 parameters that we have designed now estimated from the design sorry 3.64 sorry this is 3.3 inches not 3.64 that was the calculation instead. So, if I draw it to the scale roughly about 5 inch 7 inch then it goes down about 2 inches this is your rocket with the propellant that we have about this size rocket will produce 1000 pound of thrust with the given flow rates and everything. So, that is a small rocket which will produce the thrust actually this kind of rocket can be designed in a laboratory and can be tested. So, that is one thing I wanted to show that how we design a rocket I would like to point out here once again that the nozzle we have considered only the throat area and the exit area we have not considered the shape of the rocket that will come later when we go into more detailed of rockets now rocket nozzle. So, that is what we are going to take up next a design of rocket nozzles before that we will first define the nozzle the efficiency parameters then we go to the design of rocket nozzles and then we see how this rocket shape is designed once that is done then we go back to the combustion chamber and we will do the thermo chemical analysis to get the composition to get the temperature etcetera I will discuss how those things are obtained. So, the next topic is continuation of our discussion on nozzles and now we go into the shape nozzles. So, I will stop here today in the next lecture I will start from there.