 Welcome back to this course on rocket propulsion. So, if you recall in the last class, we have been discussing the design of rocket combustion chamber and we have solved a problem also on rocket combustion chamber design. Now, to continue our discussion on the chemical rockets, we had discussed various flow regimes for a nozzle flow. We will continue with the nozzle flow. So, first thing what we are going to do today is define some efficiency parameters for the nozzles. So, let us define some efficiencies for the nozzle. There are several efficiency parameters which are often used to describe a rocket performance. One such parameter is called cycle efficiency which is designated by eta C y. Cycle efficiency is defined as the ratio of ideal mechanical energy of the exhaust that is the hot gases that is leaving the rocket divided and the stagnation enthalpy after combustion. So, let us see what does it physically mean? The stagnation enthalpy after the combustion is the total enthalpy total energy available with the hot gases after the combustion has occurred. The cycle efficiency essentially represent out of this total enthalpy is available how much is being converted to effective thrust. So, and that is we are talking about the ideal mechanical energy of the exhaust which essentially is the kinetic energy of the exhaust. So, if I look at the rocket that we had been discussing so far. So, this is the combustion chamber and a converging diverging nozzle representing the nozzle. And after the combustion the total stagnation enthalpy in the combustion chamber is H c naught where this subscript naught represent the stagnation state and then at the exit the static enthalpy is H j. Therefore, from energy conservation or first law of thermodynamics we have shown it before that the total enthalpy here is nothing but the static enthalpy plus the kinetic energy term. So, that is H e plus now what is the kinetic energy? The exhaust velocity is u e and I represents ideal state without any losses. So, therefore, the total stagnation enthalpy after the combustion is given as H e which is the static enthalpy at the exhaust plus the kinetic energy of the exhaust. So, therefore, the cycle efficiency now can be defined as this divided by this. So, that is half u e i square by H c naught where as I have been mentioning again and again u e i is the ideal exhaust velocity. Now, first of all let us recapitulate what we have said as the ideal rocket. We have said that the ideal rocket is the rocket for which there is no pressure forces. So, the expansion is ideal. So, therefore, for ideal exhaust p e is equal to p a the exit pressure of the at the exit of the nozzle is equal to the ambient pressure that is the ideal rocket for us. At the same time we have also said that there are no frictional losses isentropic flow isentropic flow in the nozzle. Let us first understand the physical significance of these assumptions. Even if the flow let us say we have seen the different cases for at different back pressure there are different type of flows that exist in the nozzle. Let us for the timing assume that our back pressure is such that there is a shock wave inside the nozzle. If that happens across the shock wave the flow becomes subsonic and then the subsonic flow will reach ambient pressure. So, therefore, this condition will be satisfied, but is it ideal? No, because the flow is not isentropic. If you have the shock wave sitting in between the flow is no longer isentropic because shock wave is a irreversible irreversible process. So, therefore, in order to assume the ideal expansion we have to assume that the flow is isentropic. So, that there are no shock wave actually inside and then apart from that. So, there will be a shock induced change irreversibility also in the presence of friction there are some frictional losses. So, that also we are neglecting. So, with this assumption then the velocity that we will get is a theoretical velocity which we have already discussed in detail. So, therefore, that is our ideal exhaust velocity. So, now the cycle efficiency is defined with respect to that ideal kinetic energy and the stagnation enthalpy at the exit of the combustion chamber or at the inlet of the nozzle. So, now if I combined this two what we can see here is our half u e i square is nothing, but the difference in enthalpy between the inlet and exit of the nozzle because at the inlet of the nozzle the enthalpy is the stagnation enthalpy because the flow is stagnant here. So, therefore, the inlet of the nozzle the enthalpy is stagnation enthalpy and the exit of the nozzle we have the static enthalpy h e. So, therefore, the kinetic energy for the ideal expansion case is nothing, but the difference between these two enthalpies. So, we can write this expression then the cycle efficiency as h c naught minus h e by h c naught. Now, we can write this as 1 minus h e by h c naught. Now, the walking fluid at remember at the beginning of our discussion on chemical rockets we made a list of assumptions. One of the assumption is regarding the state of the walking fluid we assume the walking fluid to be perfect gas. That means, it is calorically perfect as well as thermally perfect. So, therefore, the enthalpy can be expressed as a function of temperature only. So, if we do that this is equal to 1 minus c p t e i this is the ideal temperature divided by c p t c naught where c p is the specific heat at constant pressure. Now, we have also assumed that the composition does not change in the nozzle. We have assumed that it is a frozen flow therefore, and we have assumed also that the c p is constant because temperature variation is not much there is no recombination no dissociation of the gases. So, therefore, c p is constant through the nozzle that means, the value of c p at the inlet and exit as same. Therefore, this c p let me put it as i and o just to differentiate between the exit and inlet sorry e and o let me put it like this. Now, these two are same. So, I can cancel this off. So, my cycle efficiency now is nothing but 1 minus t e i by t c naught. Now, this is the temperature ratio now and we are assuming the process to be isentropic that assumption is already there. So, therefore, the temperature ratio can be expressed in terms of pressure ratio which for the isentropic flow. Since the flow is isentropic we had earlier discussed the isentropic relationship that t e i by t c naught equal to p e i by p c naught to the power gamma minus 1 by gamma where gamma is the ratio of specific heats which we have discussed. Here p e i is the exit pressure for the ideal case and p c naught is the stagnation pressure in the combustion chamber. So, therefore, this is how for an isentropic flow we can describe the temperature ratio. So, now we can put it back into the efficiency definition this becomes 1 minus p e i by p c naught to the power gamma minus 1 by gamma. Now, p e i is a function of the shape, but we are considering ideal expansion. So, what we have already assumed that p e is equal to p a and ambient pressure is something that we know. So, therefore, here in this expression we can replace p e i by the ambient pressure p a. So, therefore, this can be written as 1 minus p a upon p c naught to the power gamma minus 1 by gamma. So, now this is our definition of efficiency cycle efficiency as we can see that this is the function of the pressure ratio nothing else. This cycle efficiency physically is the measure of the fraction of energy produced by combustion in thrust generation. The total energy is produced essentially is this the stagnation enthalpy out of this fraction how much is used for thrust generation. So, that is what our cycle efficiency represent and we have got an expression for the cycle efficiency. So, as a designer what we want to do is always maximize cycle efficiency. How do we maximize the cycle efficiency by increasing p c naught that is why for rocket application you always want to get as high pressure as possible in the combustion chamber. So, that the cycle efficiency will be more, but cycle efficiency like in any thermodynamic cycle cannot be more than 1 that is for pressure. So, therefore, this cycle efficiency cannot be more than 1 in the limiting case when we have absolute vacuum p a is tending to 0 then this ratio will tend to 0 cycle efficiency will tend to be 1. So, that is the ultimate thrust that we have already discussed. So, in the case of ultimate thrust the cycle efficiency will be leading towards 1 it will not be 1 there will be some references, but it will be very close to 1. So, either as so therefore, it says that as we go up as the rocket goes up the cycle efficiency increases because the ambient pressure continuously decreases at the rocket is launched higher up. So, therefore, this is how we define the cycle efficiency. Now, cycle efficiency as we can see is the efficiency of the entire rocket cycle which includes the combustion chamber as well. Now, let us define another efficiency parameter which is called nozzle efficiency. If you look at the cycle efficiency we have considered the nozzle to be ideally expanded, but in practical cases we will not have ideal expansion that is for sure because we have seen that there is only one case out of all that ambient pressure there is only one solution which will give us the isentropic flow. So, that is very difficult to achieve in practical big cases because we do not we would not be continuously varying p c naught. So, then p a is varying continuously. So, therefore, it is almost impossible to have a continuous application with ideal nozzle. So, therefore, the next step that comes in is to define the efficiency of the nozzle. So, when we say the efficiency of the nozzle we do not make these assumptions we say the nozzle is not ideal. So, if the nozzle is not ideal how much fraction is getting converted that is the nozzle efficiency. So, next let us define the nozzle efficiency. So, our nozzle efficiency is designated by eta n is defined as actual mechanical energy of exhaust by ideal mechanical energy of exhaust. So, now, in order to define this actual mechanical energy what is the actual mechanical energy we will represented by here our p is not equal to p a. So, if you recall at the beginning we have defined the thrust or specific impulse we have defined an equivalent velocity. So, that contains the information regarding the non idealness of the expansion. So, therefore, the kinetic energy that will be there if the expansion is not ideal will be represented by that equivalent velocity. So, the actual mechanical energy then will be half u equivalent to equivalent square. So, this is the kinetic energy at the of the exhaust for the actual nozzle which is not ideal and for the ideal we have already seen that this is equal to half u e square i u e i square. So, therefore, straight away we can see that the nozzle efficiency can be given as u equivalent by u e i square. So, what does nozzle efficiency now physically representing? If you look at the cycle efficiency cycle efficiency it tells us that out of the total energy it is produced because of combustion how much is getting converted for an ideal expansion for the exhaust. Now, out of that is ideally supposed to have been converted how much is actually getting converted that gives the nozzle efficiency. So, therefore, nozzle efficiency is the measure of the losses occurring in the nozzle during the expansion of propellant through it. However, however, how much is the losses because of the non idealness which will be frictional losses which will be walk induced losses which will be because of over expansion under expansion everything all those things are included in the definition of nozzle efficiency. So, now, with this what we can see is that u equivalent can be written as u equivalent can be written as u equivalent square by 2 is half u i u e i square by eta n, where eta n is our nozzle efficiency. And we have defined our cycle efficiency as half u e i square by h c naught therefore, half u e i square is h c naught times cycle efficiency. Now, if I combine this and this I get an expression for the equivalent velocity in terms of the cycle efficiencies and the total energy that is available for the exhaust that is the stagnation enthalpy after the combustion. So, if I combine this two what I will get is u equivalent square by 2 equal to h c naught eta cycle upon eta nozzle right. Now, h c naught is the total stagnation enthalpy at the inlet of the nozzle. Once again the stagnation enthalpy is equal to C p t c naught and C p we have shown this before is equal to gamma r upon gamma minus 1 therefore, h c naught is gamma r upon gamma minus 1 times t c naught, where t c naught is the stagnation temperature after the combustion. So, there is a temperature that exist after the combustion has occurred. So, then and we have shown that cycle efficiency is equal to 1 minus p a upon p c naught gamma minus to the power gamma minus 1 by gamma. Now, let us combine all this and get an expression for the equivalent velocity. So, our equivalent velocity will be equal to a c naught which is gamma r upon gamma minus 1 times t c naught times eta n times the cycle efficiency. So, this is our expression for equivalent velocity remember that our specific impulse is a function of equivalent velocity. So, now, just to recap what we have discussed we have shown that the performance of the rocket that the flight mechanics part is a function of our specific impulse. We started all our discussion on chemical rockets essentially to estimate the specific impulse for a given propellant system and the nozzle design. Now, what we have seen is that there is an expression for equivalent velocity in terms of the nozzle efficiency and the combustion chamber conditions. And of course, gamma and r also depends on the composition. So, after the combustion the composition of the propellants will dictate what will be the value of gamma and r. So, now, we are inching closer to our goal. Now, we have an expression for equivalent velocity this can be further written as 2 gamma minus 1. Gamma r upon gamma minus 1 t c naught eta n 1 minus p a upon p c naught to the power gamma minus 1 upon gamma. This is our expression for equivalent velocity and as we see that this is a function of t c naught t c naught the ambient pressure and gamma and r which are the function of the composition and eta n. eta n is essentially the rocket nozzle design because that will dictate what kind of losses will be there in the nozzle. So, this expression then gives us all the information required to estimate our specific impulse. Now, I would like to point out few more things here. Let us assume that the nozzle is not fully expanded is under expanded nozzle. And at the same time there are no losses because typically the losses are because of the boundary layer frictional losses are because of the boundary layer. The boundary layer is so thin in this cases that it is confined only very close to the wall. So, therefore, the momentum loss because of the boundary layer is insignificant compared to the total momentum. So, we can assume that the frictional losses are very very small. So, under that scenario if the frictional losses are very very small we can still have under expansion and at the same time there are no losses. What happens to the nozzle efficiency under that scenario? So, let me look at that case. First of all our eta n is equal to u equivalent square by 2 upon u e i square by 2. This is the ideal expansion case. Now, we are assuming that there are no losses. On top of that if the expansion was ideal if the expansion was ideal then this value then eta n will be equal to 1. So, therefore, this value is equal to this value. So, what is our ideal expansion that P e is equal to P a. So, therefore, this number u e i square by 2 essentially is the exit velocity without any losses and ideal expansion. So, that is essentially u equivalent square by 2 when the expansion is ideal P equal to P a and no losses. So, therefore, my nozzle efficiency definition then is equal to the actual case u equivalent square by 2 divided by u equivalent square by 2 for ideal expansion which is P equal to P a and no losses. So, I can write the nozzle efficiency like this. Now, what is equivalent velocity? This is the expression for the thrust f equal to m dot by u equivalent that we have derived at the beginning of our discussion. So, from here u equivalent is nothing but f upon m dot where f is the total thrust that is produced and m dot is the mass flow rate of the propellant. So, with this definition let us go back to this and put this here. So, then nozzle efficiency becomes f upon m dot for the actual case divided up by f upon m dot for ideal expansion with no losses. Remember that we have discussed the thrust coefficient, we have discussed the thrust. When we discuss the thrust we have shown both mathematically as well as with some arguments that the thrust is maximum when the expansion is ideal and for all those cases we are assume there were no losses. So, therefore, the denominator here corresponds to the maximum thrust. So, maximum thrust also means maximum thrust coefficient. So, therefore, and also this term will be proportional to the thrust coefficient. So, therefore, this is the actual thrust coefficient divided by maximum thrust coefficient. So, we can write this as C f upon C f max and it will be squared because thrust coefficient is proportional to u equivalent. So, therefore, the value of nozzle efficiency remember all this we are doing with no losses, no frictional losses. So, for no losses the nozzle efficiency is given by C f upon C f max squared this is for no losses. So, once again the value of C f max is the thrust coefficient when P is equal to P a otherwise it is C f. So, therefore, we have defined now the nozzle efficiency when there are no losses. I would like to point out here one more very critical thing this is the maximum thrust that is produced for a given ambient pressure. It is not the ultimate thrust produced by the rocket because ultimate thrust will be produced when my cycle efficiency is 1 and cycle efficiency is 1 when this term goes to 0 which means P a tends to 0. So, that is the vacuum. So, therefore, the C f max is not my ultimate thrust. So, let me just rephrase it here in this case C f max is not equal to C f ultimate because ultimate thrust is the condition when P a tends to 0. Or P c not tends to infinity, but that is highly unlikely. So, we will take P a tends to 0. So, we are going to almost vacuum then the thrust produces the ultimate thrust and that is not equal to the maximum thrust we are discussing here. Maximum thrust is we have shown that when P e is equal to P a when the expansion is ideal optimum expansion it is neither over expanded nor under expanded. So, this is the definition of nozzle efficiency. All this definitions we have been deriving so far for the nozzle efficiencies were considering converging diverging nozzles. So, this is all for converging diverging. What if we have just a converging nozzle? So, the next case we will be discussing is just for a converging nozzle. And that will bring out the efficacy of providing the diverging section once we define the discuss the converging nozzle alone. So, for a converging nozzle essentially what is the converging nozzle that a nozzle which terminates at the throat it does not have the diverging portion otherwise it would have been like this, but the diverging portion is not there it is terminated here. So, then the exit area here is my throat area exit area is equal to the throat area. Now, under that scenario what is the Mach number at the exit is 1. So, the exit Mach number for a converging nozzle is equal to 1. Now, we have shown that for the isentropic process or isentropic flow the pressure relationship as a function of Mach number. So, when we put the Mach number equal to 1 that essentially corresponds to an unique exit pressure and that will be of course, depending on the P c naught. So, for this case P e upon P c naught is equal to a function of only gamma that we have shown before. So, P e by P c naught is equal to 2 upon gamma plus 1 to raise to the power gamma upon gamma minus 1. With this then the thrust coefficient converging for just the converging nozzle will reduce to it can look at the expression for the thrust coefficient will be gamma square upon 2 upon gamma plus 1 plus 2 upon gamma plus 1 to the power gamma upon gamma minus 1 minus P a upon P c naught. Let me explain this expression. We had derived the expression for thrust coefficient before. Now, we are considering only a converging nozzle. So, for the converging nozzle the pressure ratio this is the exit pressure which now we are saying that it is not equal to the atmospheric pressure or ambient pressure because we do not have an ideal expansion in this case let us say. The exit pressure is a function of only gamma and there will be now because the expansion is not ideal there is a pressure term appearing here. If you look at this too this is nothing but this right. So, this is nothing but P e minus P a upon P c naught. So, it is coming from there and this term is to the mass flow rate m dot and u equivalent. So, that will come like this. So, bottom line is if I look at this expression for the only converging nozzle then my thrust coefficient is function of gamma and this pressure ratio P a by P c naught because my P e by P c naught is function of only gamma in this case. So, the thrust coefficient for the converging nozzle is given like this. Now, let us define some more parameters. All these parameters are actually used for design. Let us look at another parameter which is the ratio of thrust for actual nozzle which is converging diverging. So, a converging diverging nozzle divided by a corresponding converging nozzle that is we have cut that nozzle at the throat. So, the ratio of thrust produced by these two will be nothing but the thrust coefficient for the actual nozzle divided by the thrust coefficient for the corresponding converging nozzle. Now, the thrust coefficient for the actual nozzle will be we have shown this before function of gamma, the composition of the propellants, the exit pressure and the ambient pressure. So, the nozzle design as well as the composition. So, we have shown this before that this is a function of gamma P e and P naught. On the other hand for the converging nozzle alone as we are showing here that the dependence on P e by P c naught is replaced by a dependence only on gamma. So, therefore, for the converging nozzle alone thrust coefficient is function of only gamma and sorry this is P a gamma and P c naught. Let me write it as ratios. So, that P c naught also comes into the picture. So, now for the converging nozzle it is no longer the function of the exit pressure, because exit pressure is already predetermined, because we are saying that the exit the Mach number is 1. So, the exit pressure is already predetermined from this relationship. So, therefore, this is the functional relationship. If I now combine these two and look back at this ratio f upon f converging, then that becomes a function of gamma P e by P c naught and P a by P c naught the pressure ratios as well as the composition. Let us now look at the physical significance of this. For that what I will do is I will plot now one point I have to point out here that this pressure ratio P a by P c naught on which parameter does it depend? It depends on the area ratio exit to the throat area ratio. So, what will be the exit pressure will be dictated by the nozzle design, which is the area ratio is one of the prime factors. Of course, if you have a shock wave it is going to change, but here we are assuming there is no shock wave is isentropic flow there is a function of only the area ratio. So, therefore, now if I make a plot of area ratio versus this is my a e by a star a is the exit area of the nozzle a star is the throat area versus this thrust ratio f upon f converging. First of all for the converging nozzle my a e is equal to a star right. So, when this ratio is equal to 1, then this is equal to 1 also. So, let me say that I start from here 1 1. Second point we have proved that now if in order to get the expansion the area must increase that we have shown before. So, therefore, now this ratio cannot be less than 1 it will always increase. So, the ideal converging case is this right I will just draw a drop a line here representing the ratio to be equal to 1. Now, first of all let us look at this dependence p upon p c naught. So, for the given value of a by a star as we decrease p a by p c naught what will happen? This thrust is going to increase right and it will monotonically increase till it reaches the ultimate thrust right. So, therefore, if I plot this for different values it goes like this. And this is the ultimate thrust let us say where p a by p c naught is equal to 0. Remember that we have shown that ultimate thrust is also finite because that becomes a function of gamma only we do not have infinite thrust that is why it is monotonically we will approach this value. Now, on the other hand when we look at the variation with respect to p a by p c a p e by p c naught that is a function of a by a star right. So, as a by a star increases initially the thrust is going to increase reaches a maximum for the optimum and then drops that we have shown that the thrust is going to be maximum for ideal expansion. So, till the ideal expansion it will increase then start to drop. So, let me now just maintain this one and we remove this. So, now if I draw the variation with respect to p e by p c naught this is for a given value of p e by p c naught right then we increase it goes like this. So, this is our increasing p e by p c naught. What we do is we draw the variation with respect to p e by p c here that every one of them has a maximum point. So, I can draw a line joining this maximum points. So, this is the maximum thrust line this is the maximum thrust line that is the optimum thrust line. So, now I have the maximum thrust line represented here. Let us say this is we are doing for particular gamma value of 1.2. So, what we are seeing here is that this is my converging portion that is limiting here. Now the diverging portion starts diverging portion depending on this area ratio and the atmospheric pressure of course, gives different contribution to my actually this is p a by p c naught this is p a by p c naught. Let me put it p e only considering expansion to be proper. So, the diverting portion gives a contribution to the thrust in a non-linear manner initially it increases which is the optimum value when the expansion is ideal and then it starts to drop. So, for each value of p a by p c naught as we can see here this is for different values of p a by p c naught. For each values of p a by p c naught we have an optimum point and this optimum point corresponds to a specific area ratio. Now this area ratio then is our ideal expansion area ratio. So, for when we are getting this maximum at this point our exit pressure p e is equal to my this pressure. So, therefore, this ratio is equal to the ratio that we are getting here. So, for a given value of p a by p c naught there is an optimum value of area ratio which will give us the maximum thrust because that gives us the ideal expansion. Thust now what we are seeing is that if we add an area larger than this what we have is under expansion sorry over expansion and over expansion is going to reduce our thrust. So, there is an optimum area if you add area more than that the thrust is going to reduce. So, that is proved here. Now for a reasonable pressure ratio let us say if p a by p c naught is greater than 0.02 this is the reasonable pressure ratio we are considering. It can be shown that for this type of pressure ratio this reduction in thrust will not happen unless a e by a star is greater than 30. So, for this pressure ratio ambient to total when area ratio becomes greater than 30 only then the thrust is going to reduce. So, I can show it here let us say this is 30 this is 0.02. So, this line here corresponds to this pressure ratio p a by p c naught corresponding to 0.02. This is the thrust that a converging nozzle had produced would have produced. Now what we are seeing is that if we are adding the diverging portion beyond a certain area ratio the thrust produces actually less than the converging nozzle thrust. So, therefore, that ratio is about 30. So, beyond that value adding the diverging portion does not benefit at all. Before this value even though there is a reduction in thrust it is not the optimum value, but still it is greater than the converging thrust. So, we are having some advantage, but beyond this there is no advantage this is a disadvantage the thrust is reducing at the same time by increasing the area here increasing the weight. So, there is a optimum value beyond which there is no advantage optimum value of area ratio beyond which there is no advantage in adding the diverging portion less than that even though the thrust is not maximum it is still advantages over the converging portion. Now for very small pressure ratios let us say if the pressure ratio is close to 0.001 then as we can see that as we are because this is pressure ratio increasing. So, decreasing will be in this direction if you are very small pressure ratio this area ratio is very large right. So, we can have huge areas or very large areas associated with the maximum thrust. So, for the very small pressure ratios we need to have large nozzle exits in order to get the maximum thrust which may not be always possible. So, we may not have the maximum thrust, but we may have something still more than the converging portion. So, that is the advantage that will be obtaining. Now, let us look at another scenario as we keep on decreasing the pressure ratio what happens what happens we have discussed as we reduce the back pressure what happens shock wave forms right. So, as we keep on increasing this pressure the shock wave initially will be outside slowly it will move in. So, when the shock will stand at the exit right after that the flow becomes subsonic. So, that is the limiting point of ambient pressure or limiting point of area ratio we do not want to operate beyond that because the flow becomes now non isentropic there is a shock wave in the nozzle. So, there is a limiting value here this is called shock line. So, for each value of this pressure ratio there is a point beyond which we can have a shock wave going into the nozzle. So, this line represents the pressure ratio and area ratio for which a shock enters the nozzle and then we can we are absolutely do not want to operate beyond that we do not want to have a shock going into the nozzle because the flow becomes subsonic we lose huge amount of thrust right. So, we do not want to operate at all in that condition and that condition can also be derived mathematically we can derive an expression for this condition that is the pressure ratio and area ratio for which a shock will be sitting at the exit. If you increase the if the ambient pressure is little more than that the shock will go in or if the area ratio is little less than that it will more than that the shock will go in. So, this is the another limiting line which we can derive. So, what we will do now is rather I think we have exhausted the time today. So, we will stop here today in the next class we will start from deriving this expression that is the expression for the pressure ratio and area ratio for which a shock will be sitting at the exit of the nozzle. So, that will give us the shock line. So, just to recapitulate what we have discussed today we have defined this defined the nozzle efficiencies we have defined the cycle efficiency we have defined nozzle efficiency we have defined the ratio of thrust to thrust for the converging area only. We have shown that the advantage of adding the diverging area is limited to a certain range if the area ratio is more than that we do not get any advantage actually it is a disadvantages to have the diverging portion beyond a particular area ratio, but that is a function of how much pressure ratio we have and we have discussed this plot which essentially is the performance plot I would say. So, in the next class we will first discuss the relationship or the we derived the expressions when the shock will sit at the exit of the nozzle and then we will take on from there going to the shaped nozzles. So, we will stop here. Thank you.