 Till now we have discussed the examples of isentropic flows and normal shocks and we will try to utilize this understanding that we have till now to develop a feel of compressible flows through nozzles. So, we will first start with example of a converging nozzle. We have seen earlier that one of the important objectives that you have with a nozzle is to accelerate the flow and to do that let us say that you have a tank, you connect a converging nozzle with the tank. We have seen earlier that the nozzle shape always did not be convergent and in fact one could have accelerating flow with the diverging section also. So, that is a non-intuitive manifestation of the compressible flow which is not there in an incompressible flow because density variation also is very very important it is not just a into v is constant rho into a into v is a constant and rho variation is important. Now let us say that this is connected to a chamber this is the reservoir where you have roughly the stagnation conditions prevailing. Here you have the nozzle the converging nozzle this is known as a plenum chamber in which you have air maintain at the desired exit condition. The desired exit condition is not your choice but the desired pressure that is there in the chamber that you may control and there may be a discontinuity between the exit pressure and the pressure that you are having in the plenum chamber. The pressure that you are having in the plenum chamber is known as the back pressure. So, you have here the pressure as Pb which is known as the back pressure. How you control the pressure here is by regulating the flow through a valve which is located in this pipeline. So, if you want to reduce the pressure you may create vacuum by allowing more and more air from these to escape. So, what happens in this kind of a situation? Let us try to draw a graph of pressure may be we draw the graph according to the nozzle inlet and the exit. So, we draw a graph of pressure versus x. We will also try to draw a graph of mass flow rate versus the back pressure. We are plotting m dot by m dot max. m dot max is the maximum mass flow rate. So, we are normalizing the mass flow rate as a function of the Pb by P0. So, we are varying the back pressure. So, this is an experiment. Initially the pressures at the 2 places are the same. Now, we are creating lower and lower pressure by allowing the air to escape from this pipeline so that the back pressure falls. So, when the back pressure falls what will happen? So, let us say that the back pressure you have initially the back pressure falls such that it comes to a level like this. So, this is a pressure variation with x. So, when you have a low back pressure what will happen? That because of this pressure drop there will be an added mass flow rate. So, initially there is it is quite intuitive. You have initially equal pressure no mass flow rate now you have a driving pressure gradient. So, you have a lower pressure here you have a higher pressure here that will drive a mass flow rate. What do you expect? You expect that as you reduce the back pressure you will have higher and higher mass flow rate. So, if you make a plot like this so if you reduce the back pressure you will see that the mass flow rate is increasing. In this way so you reduce the back pressure you have the mass flow rate increasing and increasing till you come to a very important limit say this is the limit where the back pressure is equal to p star. What is the start condition? The start condition is the sonic condition. So, at the sonic condition you will have Mach number equal to 1. See you started with what Mach number you started with 0 Mach number so to say. So, there was no flow initially then your Mach number was increasing in this way you are coming to a state where the Mach number is 1 and there you have a flow rate which is corresponding to a velocity which is same as the sonic velocity. Now if you reduce the back pressure further so if you make a plot like this you will interestingly see that when you have a situation where pb is equal to p star you have p star by p0 then m dot equal to m dot max. So, this will become equal to 1 and then if you reduce the pressure further mass flow rate does not increase anymore. So, intuitive portion intuitive variation of the graph could be something like this. That is as you reduce the back pressure there will be more and more mass flow rate but this is not correct and you come to a saturation state which corresponds to Mach number equal to 1 beyond which further reduction of back pressure will not increase the mass flow rate. Physically why it happens? So, remember that when you reduce the back pressure what happens? It is a pressure perturbation. This pressure perturbation travels within the fluid medium at a speed of the sonic speed relative to the fluid medium it travels with the sonic velocity. So, if you write say velocity of the disturbance relative to the fluid is equal to velocity of the disturbance minus the velocity of the fluid. What is the velocity of the, what is the velocity of the disturbance relative to the fluid? It is in terms of magnitude it is the sonic speed in terms of the direction. What is the direction in which the disturbance is moving? So, disturbance is created in the downstream in terms of the back pressure. It has to propagate and go to the reservoir just in very simple terms if you have to explain it to a say junior or senior schools today it is something like this. So, you reduce the back pressure when you reduce this pressure this reservoir should know that yes the pressure is reduced. So, I have to compensate it by changing by increasing the mass flow rate. So, as if here there is a decision maker who is sitting who will know the message that there is a reduction in back pressure and accordingly will increase the mass flow rate. So, the message has to propagate. What makes the message propagate? The speed at which the disturbance propagates is governed by the sonic speed and because of the propagation of the disturbance in terms of a way through the medium that is how this reservoir knows that yes I now have to reduce I now have to increase the mass flow rate to compensate for the reduction in pressure. So, velocity of the disturbance relative to the fluid if you just take the algebraic sign of it it is minus c because let us say positive x direction is right towards the right. So, negative x direction is towards the left equal to velocity of the disturbance minus when the when the sonic condition is achieved p star then velocity of flow is c and velocity of flow is towards the right with a velocity c and this shows that you have velocity of the disturbance equal to 0. What it means is that if you reduce the back pressure further the message that such a disturbance is there is unable to propagate toward to the reservoir and make it actuated by allowing more mass flow rate. So, it is sort of say to be coming to a saturated condition for the mass flow rate and in that case the technical name is that the nozzle is choked. So, the nozzle is choked that means it has come to a critical mass flow rate beyond which if you reduce the back pressure further you will not be able to increase the mass flow rate and the corresponding Mach number at that condition is 1. If you reduce the back pressure further what will happen of course you may be able to reduce the back pressure further because reduction of pressure is your is your task that is an exercise that you are experimentally doing. So, what it will do is if you reduce the pressure further there will be a sort of oblique expansion wave outside and such that at the end. So, there will be something which is occurring in outside will not go into the details that what is occurring outside but there will be some phenomenon that is occurring just outside but important thing that we are getting out of this is that if you reduce the back pressure further at least in terms of getting the mass flow rate it will not improve any further and you can find out what is that critical mass flow rate. What is the critical mass flow rate? m dot max is equal to rho star A star u star right. What is A star? A star is the area of the exit in the converging nozzle. What is rho star? Rho star you can write p star by RT star and u star is u is mach number into C. So, m into square root of gamma RT the start condition means mach number equal to 1. So, u star is square root of gamma RT star. So, this is p star square root of gamma by R into 1 by square root of t star. You know the expression of p by p0 yes into A star. You know the expression of p by p0. You know the expression of t by t0 as a function of mach number for example t by t0 is 1 plus gamma minus 1 by 2 m square. p by p0 is t by t0 to the power oh sorry this is t0 by t it is the other way. So, this is t0 by t this is p0 by p is t0 by t to the power of gamma by gamma minus 1. So, you can calculate t0 by t star by substituting m equal to 1 and it will be a sole function of gamma. So, the expression here m dot max will be a function of what? p0 t0 gamma and A star right. So, whatever is the maximum flow rate that you will get is decided interestingly not by the pressure the back pressure which is there but the pressure at the inlet section or the pressure in the reservoir because these are the stagnation pressures and the stagnation temperature. So, this is about the converging nozzle. Let us consider what happens for a converging diverging nozzle. We can see that what is the limitation of the converging nozzle. If you want to accelerate a flow say from a subsonic state to a supersonic state that you cannot get in a converging nozzle because the maximum Mach number that you can get is 1 and that is at the exit below a critical back pressure. So, once you have that and still if you have an intention of accelerating your flow further you also have to add a diverging section and that is why the importance of a converging diverging nozzle. Before going into that maybe let us work out a problem on the converging nozzle itself. So, the problem statement is like this here in a large tank at 100 degree centigrade and 150 kilo Pascal of pressure exhaust to atmosphere through a converging nozzle with 5 centimeter square as throat area. throat area means the exit area for the converging nozzle. Compute the exit mass flow rate if the atmospheric pressure is number 1 the atmospheric pressure is like the back pressure is equal to 100 kilo Pascal. So, here there is no plenum chamber it is exiting 2 atmosphere directly. So, 1 is 100 kilo Pascal, number 2 is 60 kilo Pascal, number 3 is 30 kilo Pascal. So, clearly this corresponds to the stagnation temperature this corresponds to the stagnation pressure. So, this when this pressures are given what is your first decision making? What is the critical back pressure to get the sonic condition at the throat? So, that is the star condition. So, we know that p0 by p is equal to 1 plus gamma minus 1 by 2 m square to the power gamma by gamma minus 1. So, when you want the start condition when you want this p star then this is m star or this is equal to 1. So, from here you can find out what is p star. So, p star for this case is 79 kilo Pascal. So, 79 kilo Pascal is your p star and your given pressure is 100 kilo Pascal. So, for the one it is some state like this one say as an example which is greater than p star. So, you have not yet come up with the maximum mass flow rate. So, how do you decide that what is your m dot that will be dictated by the rho a and u which are in turn dictated by the Mach number. So, what is the Mach number? So, now you have p0 by p is equal to 1 plus gamma minus 1 by 2 m square to the power of gamma by gamma minus 1. What is for case 1? What is the value of p? 100 kilo Pascal. So, from here you can find out what is the Mach number at the exit. So, when you know the Mach number basically all the expressions are derived in such a way you know all the flow parameters. So, m dot is equal to rho exit into a exit into u exit. So, rho exit is p exit by RT exit into a exit into u exit is Mach number exit into square root of gamma RT exit and t exit by t also you can t0 by t using the expression of t0 by t is 1 plus gamma minus 1 by 2 m square from that you can find out what is t exit. If you use m exit t0 is given 293 sorry 273 plus 100 373 Kelvin. So, from here once you get t exit you may substitute all the values to get m dot and this m dot is going to be 0.15 kg per second. Now consider the cases 2 and 3. When you consider the case 2 you see that your decision making now says that the pressure is less than the critical pressure that is p star. That means the saturation in the mass flow rate has already been achieved and therefore for calculating the mass flow rate you should not go through this route but you should go through the m dot max route. So, for both 2 and 3 you have m dot equal to m dot max this is rho star a star into u star which we have just derived. So, this you can write exclusively in terms of the stagnation properties in terms of p0 and t0 and a star is the area of the exit or the throat area throat is the exit for the converging nozzle. So, if you calculate this this will come out to be 0.157 kg per second. So, you can see that this is greater than the previous mass flow rate and this is the maximum mass flow rate that you can get from this nozzle. So, with this background now we will go to the converging diverging nozzle. So, for the converging diverging nozzle let us say that you have a nozzle like this which has a converging part and a diverging part. So, if you refer to the expression for dm by m you can clearly see that for m greater than 1 dm by m is greater than 0 if dA by A is greater than 0. That means it is possible to have an increase of Mach number from the critical case which you may come here as m equal to 1. So, this part is just like a converging nozzle. So, the critical condition that you may achieve here is m equal to 1. In the converging part you cannot achieve a Mach number greater than m equal to 1, but from m equal to 1 if you want to further increase it to greater and greater and greater values then using the expression for dA by A in terms of dm by m we can see for m greater than 1 you can have dm by m greater than 0 if you have only if you have dA by A greater than 0. That does not mean that you will always be able to achieve that because that expression was derived with an isentropic flow in mind and we are not sure that in the diverging section always it will be isentropic flow. Why? Because that in the diverging section if you want to accelerate the flow you go from a subsonic to a supersonic condition and when you have a supersonic condition the flow is prone to shocks and therefore it is not trivially possible that you maintain an isentropic flow in the diverging portion of the channel. That makes the diverging portion somewhat more interesting and complicated than the converging portion. So, if you want to make a plot of again say let us do a similar experiment where we make a plot of Pb by P0 and m dot by m dot max and here we plot Pb by P0 as function of x this is the throat. So, throat for a converging diverging nozzle is the minimum area that is called as a throat. This is just a technical term. This is the throat. Initial case when no flow is like you have same uniform pressure which is Pb equal to P0 equal to 1. The ratio is 1. Now you reduce the back pressure. So, when you reduce the back pressure see there are 2 possibilities. Think of one possibility that here the Mach number is say here the Mach number is less than 1. So, the Mach number here is I mean whatever is the inlet flow from the inlet to this one the Mach number at the most may be 1 but it will in general be less than 1. So, if it is less than 1 then what will happen? So, if the back pressure is such that it could not bring it to the limit of a choking condition at the throat then what will happen? So, let us consider such cases first. So, when you have such a case you have at the throat you have what condition at the throat you have the maximum Mach number but still that Mach number is not equal to 1. It is somewhat less than 1. So, when it is less than 1 and you go along the diverging section what will happen? It will not be able to accelerate the flow. So, it will be a sort of recovery of pressure from this. So, that means if you keep this as a back pressure you may have a flow where it is accelerating in the converging section and decelerating in the diverging section just like an incompressible flow and never you have no where you have reached a sonic condition. Your Mach number everywhere is less than 1 and it is an isentropic flow throughout. So, let us say now you reduce the back pressure and similar things will occur till you come to a critical limit. So, let us say that we will consider that critical limit separately. So, first we consider the exit plane and consider the non-criticalities. Let us say that this is a state A, this is a state B. Now we come to a case where it comes to a critical state where the throat condition is Mach number equal to 1. That means the sonic or the start condition is achieved at the throat. When you have Mach number equal to 1 here what will happen in the diverging section will have 2 possibilities. Why it will have 2 possibilities? Still we are considering isentropic flow. It will have 2 possibilities with isentropic flow. It may have multiple possibilities if it is not isentropic flow but with isentropic flow it will have 2 possibilities. Why? Think about the exit area. So, you have Ae by A star. When the sonic condition is achieved at the throat then A star is equal to A throat. So, A by A star is a function of the Mach number at the exit where multiple values of Mach number at the exit will satisfy a unique A by A star. That is the equation that we have derived in the previous one of the previous lectures that A by A star is a function of it is a non-linear function of Mach number at the section that we are considering. If you are considering the exit section then the Mach number is the exit Mach number. There are 2 isentropic solutions for this. One isentropic solution is a subsonic solution. Another isentropic solution is a supersonic solution. So, if you want to have an isentropic flow at the exit with the throat Mach number equal to 1 then you have 2 possibilities. One is a subsonic flow, another is a supersonic flow. So, you have this as let us say C, let us say this is D. So, if your design is so perfect that you are keeping your back pressure same as the pressure at D what you get out of this exercise it is easy to get what is the pressure at D because you can from A by A star the supersonic solution will give what is the Mach number at the exit and corresponding to that Mach number P by P0 if you calculate that P will be the P at D which is known as the design condition. Why it is the design condition? It gives you the highest Mach number. So, it is making the flow utilize its entire potential of acceleration. So, it is accelerating to the highest Mach number with an isentropic condition throughout. So, this is known as the design condition that means of course if you keep at C still you get an isentropic flow throughout but it is a subsonic flow throughout it is not a supersonic flow. Now so if you want to utilize the diverging section with a continuously increasing velocity or the accelerating flow then the design condition is the correct condition that one may maintain but in reality it may not be possible to maintain the design condition. So, we have to see interestingly that what are the off design conditions. So, some of the off design conditions are what? So, let us say that you have a back pressure say somewhere like here say E. So, when the back pressure is maintained at E then what will happen? So, there are 2 cases the back pressure is maintained at E and the back pressure is maintained at F. So, when the back pressure is maintained at E and F. So, let us consider F as a very limiting case but let us consider first E. So, let us say that E is such that at the NC whatever is the pressure it has to match with the back pressure at the exit. So, how it may match? See the pressure was going to reduce it has to increase to match it how it may be increased by a shock. So, inside the nozzle there may be a normal shock which will make it. So, the normal shock we give symbolically like this which will make a pressure rise and match the condition with the exit condition. So, the shock will occur much ahead of the exit condition or. It depends on the back pressure at E. There were limiting condition as you reduce the back pressure that the shock occurs exactly at the exit plane. So, that is given by this F say as an example. So, here there will be a normal shock. So, if you have say or let us put a point F maybe it is difficult to put too many points in the diagram but let us just try. So, we put a point F maybe that is still a case where the shock occurs at further downstream. So, in this way as you reduce the back pressure the normal shock within the diverging portion goes closer and closer to the exit section. And if you come to a condition say G as a limiting condition there the normal shock will occur at the exit plane. So, G corresponds to normal shock at exit. If your back pressure is between G and D that also maybe a case. So, let us say that the back pressure is at H which is between G and D. Then this is accompanied by because see this discontinuity has somehow to be adjusted. So, that is adjusted by a shock not within the nozzle but just outside the nozzle and that is an oblique shock. So, this is oblique shock outside the nozzle. If the back pressure is less than D some greedy designer may think that well I want to reduce the back pressure below D and C because it might be technologically possible. So, let us say I pressure curve which yes once the back pressure is maintained at H means the oblique shock will go and mean how will meet the curve. See this oblique shock will occur just outside the exit. So, this is a symbol this is not a pressure curve this is just a symbol of oblique shock that is all. Just a symbolic way of representing it. The whole idea is that whatever maybe the pressure here the pressure has to eventually match with the back pressure. See, so when it comes to the pressure at this point say you think about this think about this point. So, when you have the pressure same as the back pressure same as G. So, till the exit it has come with an isentropic condition. It has come to a point very close to the point D. It is not technically same as the point D but infinitely similarly away from the point D just at this one but you have kept your back pressure same as the point G. So, at the exit there should be a discontinuity to match with the exit pressure from. So, the exit pressure should go through a discontinuity to match with the back pressure and that is through a normal shock at the exit. Similarly, if the back pressures are lower and say here that discontinuity is not in the form of a normal shock but in a form of oblique shock just outside the exit plane. When you come to the point I which is less than the where the back pressure is less than the point D. So, let us say that you want the your back pressure at I. When you have your so up to D very close to the D it could come to what state? It could come to a state where it is an isentropic flow almost throughout very close to the point D but back pressure has to match with pressure at I. So, there should be some sort of discontinuity between D to I that is in the form of what? See D to I is what pressure increase or pressure decrease? So, it is an expansion shock wave is the till now whatever we have considered as a shock wave is the compression wave. So, here it is an expansion. So, it is an oblique expansion wave that is there at the exit. So, this is an oblique expansion wave occurring outside the nozzle. So, if you consider from a design perspective D is the back pressure at D is the design condition. If you have your back pressure between G and D then still you get almost entirely isentropic flow within the nozzle. If you consider your back pressure less than D then also you get almost entirely isentropic flow within the nozzle. But the discontinuity to match up the back pressure and the exit pressure is through different forms of waves shock wave or expansion wave in terms of m dot by m dot max. Now, let us see this graph. So, when you reduce your back pressure just as the previous case the mass flow rate will increase. So, when the mass flow rate increases it will come to a saturation out of these all these points in the diagram which point will be the point corresponding to the saturation mass flow rate beyond which mass flow rate will not increase further C. So, C because it corresponds to the critical the sonic condition gives you the maximum mass flow rate. See for the mass flow rate converging portion is important because whatever mass flow rate the converging section is delivering that is going through the diverging section. So, you come to the point C maybe A is this and B is this then if you come to this one this is m dot by m dot max equal to 1 all your points like E, G, H, D, I will give the same maximum m dot which is given by the sonic condition at the throat. So, it means that if you want to have a supersonic flow in the diverging section it must pass through the sonic point at the throat in a converging diverging nozzle that is a very important understanding. So, we have seen that if you have such a geometry where it is a converging section and a diverging section it is not always necessary to have achieved a sonic condition at the throat. So, just in the previous class we have worked out an example where the geometry was like this and at the throat we got a condition where the Mach number is less than 1. So, it is not a must that at the throat the Mach number should be equal to 1 but if it at all has to be 1 it has to be at the throat. So, these are the conditions for which the Mach number is 1 and that is at the throat these are the conditions for the Mach number is less than 1 right. So, with this understanding now let us try to work out a couple of problems on the converging diverging nozzle. So, the problem is like this air flows through a duct as shown in the figure the duct is like whatever is drawn in the sketch with there are 3 sections section 1 with given m 1 equal to 2.5 p 1 equal to 40 kilo Pascal and t 1 is 30 degree centigrade. The throat name is section 2 no sorry throat name is not identified as any section it is told that in some section 2 a normal shock occurs wherever it is a normal shock occurs here there is a section 3 also which is Mach section 2 has a 2 is equal to 18 centimeter square section 3 has a 3 equal to 32 centimeter square and a 1 is 24 centimeter square these are given. So, with this given data you have to find out the Mach flow rate number 2 the Mach number at section 3 and the stagnation pressure at section 3. See at the section 1 everything is given to calculate the Mach flow rate because at the section 1 you have p 1 t 1 and therefore you can find out rho 1 and you have the Mach number that means you know what is the corresponding u 1. So, m dot is same as m dot 1 that is rho 1 a 1 u 1. So, rho 1 is p 1 by r t 1 a 1 is given u 1 is m 1 square root of gamma r t 1. So, all these data are given if you substitute that you will get this as 0.96 kg per second. Now you move on to section 2 see section 2 is such a section where you have a discontinuity. So, you have a 2 upstream and you have a 2 downstream the properties are different there because there is a shock at section 2. So, at the section 2 how do you find out what is the Mach number in the upstream of the section 2 you can write a 2 by a 2 star as a function of Mach number at 2 upstream when you consider a 2 star as upstream a 2 star. Remember a star changes across the shock so when we say a 2 star it has no meaning is it in the upstream side or is it in the downstream side accordingly the Mach number is upstream side or downstream side. So, how do you calculate a 2 star in the upstream it is same as a 1 star. So, because from 1 to 2 upstream is isentropic flow. So, a 1 by a 1 star is what a 1 by a 1 star is function of Mach number at 1. So, from here you can get what is a 1 star which is same as a 2 star upstream. So, you can from that you can get what is a 2 by a 2 star upstream once you get what is a 2 by a 2 star upstream this if you look into the table of the compressible flows it will have 2 solutions one is supersonic another is subsonic which one would you accept as m 2 upstream supersonic because the shock is from supersonic to subsonic that is what the second law of thermodynamics says. So, from here you get the supersonic solution for m 2 and that m 2 upstream is 2.18. So, once you know m 2 upstream you can calculate m 2 downstream because you have the formula for m 2 by m 1 for a shock. So, m 2 upstream this will give you what is m 2 downstream by m downstream by m upstream formula for shock that is m 2 square in terms of m 1 square that formula. So, that will give you m 2 downstream equal to 0.54977 when you have got an m 2 downstream you can calculate a 2 by a 2 star downstream. So, a 2 by a 2 star downstream is a function of m 2 downstream. So, from here a 2 stream remains the same so from here you can get what is a 2 star downstream and that is same as a 3 star. So, then from this you therefore have what is a 3 by a 3 star this is a function of m 3. So, out of the possible m 3 is again it has 2 solutions which m 3 we will take subsonic or supersonic because it is downstream of the shock. So, this you get what is m 3 as the subsonic solution that is equal to 0.27. So, that is the part 2 of the solution. What is the part 3 of the solution? What is p 0 3? So, how do you find out p 0 3 that is yes how do you calculate p 3? How do you calculate p 0 3? This you tell how do you calculate p 0 3? See you have p star okay. Then so you have p 0 by p as a function of what 1 plus gamma minus 1 by 2 m square to the power gamma by gamma minus 1. So, when you consider p star so what is the how do you get the corresponding p star? See remember p star is a reference state which is not any physical state out of 1, 2 or 3. So, in principle obviously if you know say p you could calculate what is p 0 but how do you calculate p 0? You need to refer it with a star state. How do you calculate that star state? Let me give you the answer exactly. If you want to calculate this in this way the trouble is you are you are required to know what is p but if you forget about that if you forget about that just go to the upstream of the shock. Can you tell what is p 0 at the upstream of the shock? Yes p 0 by p 1 you know right. So, in the upstream of the shock you know p 0 by p 1 as a function of m 1 right. So, this is p 0 at upstream right. Can you relate p 0 at upstream with p 0 at downstream? That is p 2 by p 1 that is p 2 up that is a p 2 ratio what about p 0 ratio? No no downstream once you have the the situation is from here you can find out what is p 1 sorry what is p 0 upstream. So, you can calculate p 2 upstream once you calculate p 2 upstream you can calculate p 2 downstream right. So, this will give you the possibility of calculating p 2 upstream p 2 upstream because p 2 by p 0 is a function of mach number at the upstream upstream p 2 upstream by the shock relationship normal shock relationship will enable you to calculate p 2 downstream right. When you calculate p 2 downstream. You get p 0 downstream because p 2 downstream by p 0 downstream is a function of mach number at 2 downstream. So, p 2 downstream by p 0 downstream is a function of mach number at 2 downstream which is already calculated. So, from here you can find out what is p 0 downstream that is same as p p p 0 3 okay. But, if you use the table of shocks you need not do all these it is simply given a p 0 up p 0 downstream by p 0 upstream because it goes through all these calculations and eventually tabulates in the form of that. So, if you do that this you will be 435 kilo Pascal that is the answer to the third part. So, that is the p 0 downstream okay. So, we can see that given certain conditions we need to keep in mind that a star is a reference quantity. The a star may not be the any physical a star which we are getting is waiting for these sections. It is just a hypothetical state where in an isentropic condition the system is brought to a sonic state. So, it is different for the isentropic flow in the upstream and isentropic flow in the downstream. So, what we are assuming is that upstream of the normal shock is one isentropic flow downstream of the normal shock is an isentropic flow. So, in the upstream of the normal shock p 0 t 0 a star they are remaining same downstream of the normal shock p 0 t 0 and a star they are remaining another set of same values. And there is a discontinuity of those as you go across the shock because of the non isentropic nature of the flow across the shock okay. We stop with this lecture now and we will continue with the next lecture. Thank you.