 So, in the last couple of lectures, we have been discussing shaped nozzles. We had first discussed a conical nozzle and where the nozzle is a cone. And we have shown that because of the three dimensional effect, there is a loss in the performance, which was given by a factor lambda, which depends on the half cone angle. Then, we have said that this loss is primarily because of the flow deviating from the nozzle axis because of the three dimensional effect. Then, what we said is that if we go for a shaped nozzle, if we contour the nozzle wall in such a way that the flow turns and towards the exit, it leaves parallel to the axis of the nozzle as well as we get a uniform flow, then this losses can be minimized. And based on that, we started our discussion on design of such contour nozzle or shaped nozzle. We started discussing the use of method of characteristics for nozzle design. We have shown, we have said that the method of characteristic is applicable only for supersonic flows. It cannot be used for subsonic flows and the inherent assumption in method of characteristic is that the flow is isentropic. Based on that, what we have done in the last class is, we started discussing the design of a curved nozzle using method of characteristics. We said that, let us say this is our nozzle shape roughly like this. And we have also said that the converging portion is relevant because the flow is accelerating there is a favorable flow pressure gradient and the flow is subsonic. So, there is not much a problem in designing that. So, we can just choose a contour which is rather not a contour that is a converging portion and get away with that. The diverging part however is more critical. So, we have to put in more effort in designing the diverging part. Then what we have said is that, when we start to design, let us say we choose a particular point here. So, let me once again draw the contour. Let us say this is the inlet point, this is the nozzle axis. We said that the characteristic lines are formed from here because of the slight change in the angle. If I expand this, if the supersonic flow is coming like this and then there is a slight change in angle, about this turning there is an expansion fan that is created because of that expansion fan the flow turns and becomes parallel to this side at the same time it accelerates. So, Mach number increases. So, towards the beginning section then we have what we said is that we take this section and break it into many small sections. So, that there is a gradual change in the angle and then we said that starting from both these points the characteristic lines are formed which are going to go like this and form a kind of grid. It is not exactly cutting the way I wanted to, but anyway it forms a grid like this. We also said that in this location let me call this A, this is C, this is B. Similarly, this is A prime, this is B prime, this is C. In this location, in this area rather portion of the segment if we see that we have only one sided characteristic lines. On this side it is left running, on this side it is right running. So, there is only one sided characteristics that are present. So, then we had estimate since the variation in angle theta or rather d theta here in this case is specified we can estimate the changes in properties. So, how will we do that? Let us say if I look at this flow again we have said that this flow will be turning if the flow velocity coming here is u and then it turns becomes parallel here at the same time it gets accelerated. So, this velocity becomes u plus d u. Now, initial velocity was u. So, there is an increase in velocity d u and this d u can be split into two components. One is the horizontal component d u other is the normal component d v and so therefore, this is our d u. Based on that then we use the definition for the Mach angle which was tan alpha is equal to d u by d v and at the same time tan alpha equal to 1 upon m square minus 1. This is the definition of Mach lines and this is the Mach angle from there we can we have. So, using this definitions as well as the definition of speed of sound or the Mach number and isentropic relations we had derived relationships for the change in velocity, change in Mach number, change in pressure, change in temperature and change in density. So, this is what we have done in the last class. So, by doing so what we have done is till point b this entire zone here all the conditions now are known till starting from point a to point b all the conditions here are known and here the Mach lines are not crossing each other. Now, the next thing that we have to discuss is if I look at this zone the Mach lines are crossing each other if they are crossing and if I look at this after every crossing there is a change in property. So, now everywhere there is a change in the properties like this of course, this will not be the same equations or same expressions, but the properties are going to change. So, now what we want to do is let us discuss what happens when the Mach lines intersect each other. So, this was the recapitulation of what we have done in the previous class. Let me just remove this and focus on the intersection of Mach lines and how the properties change. So, the region downstream of this region which we have been discussing so far the Mach lines cross each other. So, therefore, conditions downstream of this Mach lines are no longer uniform which were uniform previously because Mach lines when was crossing. So, now let us look at intersection of Mach lines. Let me consider two Mach lines one left running one right running and they are crossing at this point O and after crossing the Mach lines change the deviate from their original path and go like this. So, this is the incoming Mach line A 1 and B 1 after crossing they become A 2 and B 2. So, now let us see what happens to the flow. Here the flow is uniform in this domain coming with a Mach number M. Now, first let us look at the bottom one. This flow comes to this Mach line when it crosses this Mach line it gets deviated and it goes so like this. Let me consider this deviation for this Mach line is D theta 1 and then after crossing the Mach line not only it gets deviated it gets accelerated. So, the Mach number is going to change. So, in this branch let us say the Mach number is let me call this M 1 or M only so M plus delta M 1. So, initially this was M now it is incremented by the amount delta M 1. After that this Mach line this flow crosses another Mach line which is O B 2. So, when it crosses this Mach line O B 2 there is going to be another change again it will get deflected at the same time the flow will get accelerated. So, let me this was the initial path. If I extend this this is the initial path of this flow coming into this Mach line now it gets deflected say like this. Let me say this angle is D theta 2. Now the flow here after it has crossed this O B 2 line has further accelerated. So, initially now when this was M plus delta M 1 now it becomes M plus delta M 1 plus another acceleration delta M 2. Now this angle that is the see this is the flow direction for this Mach number Mach number and this is the Mach line. So, therefore this angle alpha 2 is the Mach angle for this Mach line with respect to the incoming flow. Same thing happens for this sum. So, let me do it here first it gets deflected by an amount let us say D theta 3 and gets accelerated D M 3. Then coming here it is further changing. So, this was the original direction now it changes like this and this deviation is D theta 4 and this angle is alpha 4 and the Mach number after this deflection is. So, essentially what happens is that in this region there is no Mach line. So, therefore this 2 flow should be same we should know we do not want to have a difference in Mach number. So, initially it comes with a given Mach number. Then this Mach line and this Mach line may have different strength and different angles. So, because of that there are deflections may be different, but finally what we get is this. So, the flow condition upstream of this line O A 1 which is this side as we can see here. The flow condition upstream of this is uniform, but upstream of this if I look at O A 2 there is a change between this and this. So, therefore this Mach line and this Mach line are different and the flow conditions are different between this region and this region. So, because of this the Mach lines are changing the directions the Mach line is changing the direction. It is not going straight in this direction because the flow conditions are different because this Mach line is determined by this Mach number. This Mach line is determined by this Mach number and since this and this are two different Mach number Mach lines deflect or deviate from the original path. Now, if we know the flow in this region this as well as in this region A 1 O B 2. So, if you know the flow between B 1 O A 1 A 2 and A 1 O B 2 then we can determine the flow in this region also that is the point. That is if we know this change that is occurring to the flow we can get what is going to be the conditions here. So, that is what the bottom line is of all this analysis that how do we get this changes. So, the deflection of the stream line occurring in this O A 2 and O B 2 these deflections are not arbitrary they are actually determined by this right. Let me write this expression one this we had derived in the last class. What does this say that the change in Mach number depending on a change in theta right. So, if the theta changes the Mach number changes. Now, if I look at this region itself what this region is seeing essentially is the same thing which we have done there that a supersonic flow crossing a Mach line right nothing else. So, therefore, this is applicable to the flow here. Similarly, it is applicable to the flow here. So, therefore, what will happen to the flow here is dictated by that expression that we have a change in Mach number and that change in Mach number for this region is delta M 2 for this region is delta M 4 and this is not arbitrary it will be governed by this equation. And the there is a requirement that all the stream line in this region are parallel. So, after turning the flow should become parallel here. So, flow should be going in one direction that is what we want to get right we want to get a parallel flow. So, let us say that we design in such a way that the flow becomes parallel. So, we have a parallel flow in this region in A 2 O B 2 at the same time since we are having a parallel flow we should have same Mach number also once again that is the requirement we want to have uniform parallel flow right. So, therefore, in this region where there are no Mach lines. So, just to refresh what we have been saying when the region where there are no Mach lines if the flow is not crossing any Mach lines or the region between different crossing and different Mach lines the Mach number should be same uniform flow at the same time the flow should be parallel. We do not want to have because if the Mach number is different we have different Mach lines right. Similarly, if the flow is not parallel we are going to have different Mach lines. So, therefore, we want to have parallel and uniform flow between the Mach lines. So, similarly in this region there will be parallel and uniform flow here flow is parallel and uniform here of course, we are assuming the flow is parallel and uniform right. So, every between every crossing the flow should be parallel and uniform. So, now let us look at this domain. If I look at A 2 O B 2 we want to have parallel flow what does this mean parallel flow means the flow was parallel here it was deflected twice and then it takes a final direction. Similarly, the flow was parallel here it was deflected twice and then it takes the final direction what we are saying is that this deflection for this flow is equal to the deflection for that flow. Now, how much this flow was deflected d theta 1 minus d theta 2 how much this flow was deflected d theta 3 minus d theta 4. Now, the flow was parallel here. So, in order for this to be parallel this deflection should be equal. So, therefore, we write d theta 1 minus d theta 2 is equal to d theta 3 minus d theta 4 let me call this equation 1. Similarly, we are saying that the Mach number is same here in uniform flow. That means, whatever increase in Mach number has happened from here to here is same as the increase in Mach number from here to here. What is the increase in Mach number for this branch it is delta m 1 plus delta m 2 for this it is delta m 3 plus delta m 4. So, this 2 should also be equal. So, therefore, delta m 1 plus delta m 2 equal to delta m 3 plus delta m 4 let me call this equation 2. Now, if I look back at this equation let me give you the name equation a. If I look back at this equation this gives an expression for change in Mach number with respect to theta. So, I can write this as d theta is equal to some constant m times d m d theta equal to some parameter m times d m not constant it will be function of Mach number also. So, therefore, if I use it for different paths now let us say first let us use it for this one rather for this one d theta 2. So, our d theta 2 will be equal to some parameter m 1 times delta m 2. Similarly, delta m 2 for this one d theta 4 is equal to some parameter m 3 times delta m 4 let me call this equation 3 and this is equation 4. Now, this values of m 1 and m 3 comes from this equation these are not arbitrary and those values can be obtained as putting this equation I will just write down the expressions m plus delta m 1 square minus 1 divided by m plus delta m 1 minus sorry 1 plus gamma minus 1 by 2 m plus delta m 1 square. Where is it coming from? From this equation where if I look at d theta the incoming Mach number is m plus delta m 1. So, that is what we have put there. So, we get this expression similarly we get m 3 which is appearing here is equal to square root of m plus delta m 3 square minus 1 divided by m plus delta m 3 1 plus gamma minus 1 by 2 m plus delta m 3 square. So, as we can see here m 1 and m 2 are functions of the incoming Mach number m and then the deflection or the the change in Mach number that has occurred because of the first crossing. So, I get this expressions. Now, let me look at this we have 1, 2, 3, 4, 4 equations and what are the unknowns that we have? So, we have 4 equation 1, 2, 3 and 4 and my unknowns are d theta 2 d theta 4 that is this change and this change because this is known this change is known. We know exactly because this is the single type of Mach line it is not crossing. So, we have already derived all the expressions for that we know how much theta is going to change and everything. So, this is known we are interested in this zone. So, we need to know this angle and this angle plus this 2 Mach numbers del m 4 and del m 2. So, these are the 4 unknowns that we have. Now, we can solve for this 4 equations with 4 unknowns and we can get all the values that is how much change in angle will be there that is how much flow is turning change in angle is how much flow is turning and how much change in Mach number is there. Once we know these 2 we can use the other equations equations last lecture I had listed 5 equations. We can use now those other equations and get the change in pressure density temperature velocity Mach number everything. So, as soon as we get these 2 solved other things can be obtained. So, for example, if I am interested in getting delta m 4 then from this 4 equation delta m 4 will be equal to divided by m 3 minus m 4. So, for this 4 equations this is the value of delta m 4. Once again m 1 and m 3 as you can see is function of incoming Mach number and the first change which is known. So, delta m 1 and delta m 3 are known in this equation. So, therefore, m 3 and m 1 are known. So, I can get the change in the Mach number after it crosses the second one. So, essentially if I can summarize it that if I know the incoming flow Mach number uniform Mach number if I know the initial angles d theta 1 and d theta 3. Then of course, from here I know delta m 1 and delta m 3 right using that equation I get this parameters. Once I know this then the entire flow in this region that is a 2 o b 2 is can be estimated. So, if I know the flow in this region and what is happening across this Mach lines. So, if I know the characteristics of this Mach lines for that I need to know the characteristics of this Mach lines then the entire flow can be estimated. Now, there is one glitch here if this and this Mach lines like we have done here. If I look at this Mach line and this Mach line from the first point then this and this Mach line are originating from exactly same conditions. So, therefore, this Mach line and this Mach line will bring in exactly same change. So, what is happening there in that case that if they are originating from the same condition and bringing in the same change then what happens is that d theta 1 is equal to d theta 3 right as well as del m 1 is equal to del m 3. Now, if del m 1 is equal to del m 3 then from here what we see is that m 1 is equal to m 3. So, therefore, m 1 is equal to m 3 now under such scenario this is 0. So, this is indeterminate. So, under such case del m 4 is indeterminate then how do we get this how do we solve for this if they are coming from the same and similar thing will happen here not only in this zone. Let me draw this here in this zone right where the Mach lines are crossing and then we are getting into this condition. How do we get this thing the what we will do is under such scenario we know that the flow is parallel here flow is parallel here. So, if the flow is parallel here it is parallel to this here also is a parallel flow. So, therefore, the flow in this region is parallel to the incoming flow. So, because of that since the flow is parallel to the incoming flow this value d theta 4 is known because the final turning is known how much it is turning at the final condition. So, d theta 4 is known in that such scenario then we do not have to use this equation to calculate instead what we can do is we use there is a fourth equation in our list right. So, if d theta 4 is known we can just use this equation to calculate the change in Mach number we do not have to solve all four together. So, essentially that simplifies that if we know for example, for this case that the flow is originating this two Mach lines are same then straight away we use this, but it is not applicable to flows here because here this Mach line is different this Mach line is different. So, for this zone I have to use solve the four equations, but only here I use this to get this condition. So, therefore, going doing like this we can solve for the flow downstream of this condition also now. So, this was our initial domain where all the Mach lines as we have seen were same type. So, this was a b c a prime b prime. So, up to this we have seen that the deflection is known now beyond this beyond this domain b c b prime here the Mach lines are crossing right. So, now we can solve for theta also earlier our this theta was specified now we are solving for theta. So, when we solve here now this also emerges as a solution right this also emerges as a solution. So, we now can calculate the flow downstream of this crossing point downstream of c after that we can calculate downstream of each of this point def etcetera. So, now we can get the full flow domain solution we know now the pressure everywhere pressure Mach number density velocity temperature everything can be obtained. So, what we have seen now is that initially this has to be specified after that we can design the flow in such a way that we get a uniform flow at the exit. Let me now just list the procedure that is to be used to do that. So, first of all we need to know these conditions not only that this you have taken a this is essentially making a computational grid we can also do is we can take conditions in between also and make a final grid resolution will be better. So, we can have conditions here from here the Mach lines will emanate etcetera so now let us say for the time being we are just focusing on this. So, we need to know the Mach number here and here also that is my inlet condition. Now, there is a problem here from where we should start should we start from the throat should we start before the throat should we start after the throat. If we start before the throat what is the Mach number there is less than 1. So, method of cactus is not applicable if we start at throat Mach number is 1. So, therefore, what is alpha 90 degree again that cannot be used. So, we have to start slightly ahead of it slightly ahead of the throat instead of starting from slightly ahead with a length what we say is that we start from a Mach number slightly higher than 1. So, the standard practice is to start at a location where Mach number is 1.1. So, we start at a location where Mach number let us say is 1.1. So, this is now our assumed initial point the number of starting points now we can select at that plane. So, let me just draw a schematic here this is our initial rather the axis the converging portion throat then the diverging portion. So, this is my throat I am starting somewhere here this is my point a where Mach number is say 1.1. Now, what we do is we can choose many many points here depending on our computational facilities available how complex computation can be done. So, we choose multiple points in this line everywhere in this line Mach number is 1 sorry 1.1 is a uniform flow coming here that is what we are assuming now let us look at what is happening from here there will be a Mach line going like this from each point here there will be Mach line going like this straight right these are going to be straight lines only here it will curve. So, let me call this point I now similarly from all this points here they are going to be Mach lines coming like this similarly from different points coming from the bottom from the bottom half also there will be Mach lines which are coming in. So, now what you can see here is that we have created a grid of Mach lines starting from this point a let me say this point is b this point is c etcetera let us say this point is h this is i similarly I have j k etcetera up to p. So, now we have created the entire Mach line grid where the Mach lines are starting from different locations. So, now we start our calculations from here and proceed downstream note that the flow is always along the stream lines right. So, essentially we are getting the solution for the stream lines and since the flow is along the stream line we have one stream line at the center, but the flow is like this right. So, this direction we know another thing we know is the initial curvature this wall is also a stream line right. So, we know the flow is going to be along the wall and say initial this information is known. Now, after making this grid following this method of characteristics we can consider the flow to be isentropic we can get that 2 D flow field right this is a 2 D method that I have already said. So, the two dimensional flow field can be estimated completely by following this method and that is how we design a perfect nozzle. So, what we are doing is we start from this line initial point and start to solve this problem. At some point here I when we reach the flow field I let us say at this. So, now when the flow is going here it gets accelerated right. So, at some point say at I we say that we get the design Mach number. So, as we marching our solution we get the design Mach number at point I. Now, we know that we have reached a condition where we have our design conditions. Then this is the exit Mach number design exit Mach number I like to point out here that this design Mach number may not exactly coincide with the point of intersection it may be in between two or in between the two intersections. So, then in that case we have to interpolate we have to interpolate and get the design Mach number. Now downstream of this I once we have reached this after that this Mach cone that is emitting from here let me draw the Mach cone here this Mach cone that is emitting from this now we have to ensure that everywhere here now the there is no more Mach lines coming to this zone right. So, we have identified this now we have drawn a Mach cone if there is no more Mach line this entire domain here the flow will be parallel and uniform right that is what we want because we have shown that once we cross the crossing of two Mach lines as a domain after that the flow is going to be parallel and uniform. So, we are going to have Mach number equal to m e and it is parallel flow. So, now we have identified the exit Mach number and now we know up to what point we have to extend our solution domain. So, the downstream of this point I the correspond and the in the corresponding Mach cone the velocity is constant and the flow is parallel to the axis. We can now draw this Mach line I p as shown here at an angle alpha e right and that depends on the Mach number that we have got here m e right. So, we can draw this Mach cone alpha e now once we have this then along this line every point here the properties are known because they depend on the Mach number right. So, the properties at this Mach line is known Mach line I p are all known now. So, that is how we get up to the exit point that is for the time being just take a step back here how do we reach this contour or attain this contour for that now let us look at this wall initially this wall curvature must be provided like we have already discussed initial curvature for the wall must be provided, but after we are out of that influence zone where we have only a single type of Mach lines after that the solution will tell us what should be the wall contour. So, let us say downstream of this point the wall contour is no longer arbitrary because as we can see here there are multiple Mach lines crossing right. So, that is no longer arbitrary arbitrary contour will be chosen between these two points after that now we do not have arbitrary contour. Now, the steam lines in this region which we are having here that will be determined by the known properties at all this points a b c d etcetera all this points the properties are known. If I look at now this let us say let me remove this if I look at this let us consider this is a Mach line with a stream line we have two Mach lines crossing here right this and this. So, we can estimate the properties here right and we can estimate how much curvature will be provided based on the analysis we have just done right. So, we know how much turning needs to be provided then we solve for this let us come to this position now. So, now here again we have one Mach line here one Mach line here we can estimate what will be the condition. So, like that close to the wall we have like this. So, depending on this we calculate for this location then depending on this we calculate for this location like that. So, we progress calculating the wall contour d theta all along from point here to point here I to a to p that is how for every point here we estimate how theta should be varied. Then finally, we get this contour and at the end what we say is that now we stop here this is our nozzle exit everywhere here there is a contour the flow stream lines are parallel to this wall and finally, here we get a uniform parallel flow. So, from the known values now which we are calculating from different location we can fill in the solution between this entire domain from here to here, here to here and everything. Once this is obtained the solution is obtained now we can draw the stream lines also we can draw the stream lines and we can get the wall contour. The stream line between a and p is my wall contour actually right. So, if we can once we solve for this we know how much the flow is turning what are the flow properties. So, this solution will give us all the properties it will give us the wall contour it will give us the Mach number variation it will give us the pressure variation density variation temperature variation everything in one go as a package that is something that is a very useful information. Now, the accuracy of course of how accurately it is predicting depends on how many points we choose here. So, how fine is our mesh finer the mesh better will be the accuracy, but then beyond a certain number of mesh the incremental advantage is going to diminish therefore, we do not want to go to when you find mesh because as we increase the mesh the computational time increases. So, we do not want to go to very fine mesh. One thing like to point out here that if I look at this solution procedure we are not solving any differential equation anywhere right essentially the equations we are solving are all algebraic equations. So, that is a very simple technique. So, therefore, there is no finite differencing or anything required it is just solving set of algebraic equations again and again. Therefore, the solution is also very fast and that is one of the major advantage of this method. We do not have to solve full may be a stokes thing is that. However, there are some assumptions that we have made which are going to be problematic. For example, we have assumed that the flow is slipping this is the streamline the flow is slipping right. We have assumed that the velocity at the wall is the slip velocity this is something that is not going to happen in practical cases. We are going to have no slip boundary condition there in practical cases there will be a boundary where the velocity is 0 or then slightly increases up in the boundary layer right that we have not accounted for not only that close to the wall in reality even if the flow here is supersonic we have subsonic flow close to the wall that is something that is not accounted for here and actually this method will not be able to even capture that because the flow becomes subsonic the method is going to collapse right. So, therefore, this is applicable only in the supersonic domain that we have to understand and not only a supersonic domain it is completely outside the boundary layer where the flow is essentially not affected by the boundary layer at all. However, by using this method we get the contour we get the length also as you can see from here to here we get the length of the nozzle also. So, the as a initial estimate this is a very good technique right. So, we can get the undisturbed or free stream flow without the boundary frictional effects we can get the contour we can get the length later on we can correct it by solving full navier stokes, but then the design is there initially I said in the last class that for solving full navier stokes you need to have this geometry. Now, we have the initial geometry which has been obtained from a simple method now we can correct our predictions for this geometry by incorporating the real flow effects. So, what I will do is I will stop here now in the next lecture I will talk about this real flow effects and how do they affect the performance of the nozzle. Thank you.