 So, let us work out an example involving the concepts that we discussed in the previous class. A converging diverging nozzle with an exit to throat area ratio of 3.5 operates with inlet stagnation conditions of 1 mega Pascal and 500 Kelvin determine the exit conditions when the back pressure is 20 kilo Pascal A and B 500 kilo Pascal assume air to be the working fluid, okay. So, we are given the inlet stagnation conditions. So, in the analysis of such problems what we would typically do is determine the exit pressure corresponding to solution F, solution E and solution B. So, once we do this then we can actually determine where the given exit condition lies and then proceed accordingly, okay. What is that B and C? So, solution B and I am sorry B and F are actually isentropic solutions for the given area ratio. So, isentropic solution for the given A exit over A throat. So, B corresponds to the subsonic solution. So, B corresponds to M exit being less than 1 and F corresponds to the supersonic solution with M exit being greater than 1 but same area ratio isentropic solution, okay. So, let us proceed. So, let us first retrieve the value from the isentropic table corresponding to supersonic exit. So, for the given area ratio which is 3.5 we may retrieve the following quantities. So, 3.5 and supersonic solution so A over A star 3.5 lies here, okay. So, Mach number at exit is 2.8 and P not over P, T not over T or all may be retrieved from the table, okay. So, exit Mach number is 2.8, P not over PE may be retrieved from the table. So, the exit pressure for this case is 36.85 kilopascal, okay. So, if you go to the previous one instead of plotting P0, P over P0 inlet if you plot, okay let us try to plot it afresh, okay. So, throat let us say this is the exit. So, the supersonic solution the pressure is 36.85 kPa. So, this corresponds to solution F from the previous graph. Now, supersonic exit again for the same area ratio we now retrieve the supersonic solution. So, for the same area ratio we go to the supersonic branch. So, 2.5 area ratio falls between these 2 entries. So, Mach number falls between these 2 values and P not over PE, okay P not over PE falls between these 2 values. Yeah, the ratio is 3.5. So, that means it falls between these 2 entries and Mach number falls between these 2 entries, okay. So, we have to interpolate and retrieve the values. So, if you do that the exit Mach number comes out to be 0.1682 and exit pressure comes out to be 980.4 kilopascals. So, that would be the subsonic isentropic solution as this uses slightly different color to denote that, okay. So, that corresponds to 960.4 kilopascals. So, 980.4 kPa. The next solution is the one corresponding to normal shock exit or curve E that we have shown in the previous graph. So, for this case if you recall we have the nozzle. So, we have the nozzle that looks like this and a normal shock stands just at the exit, okay. Since the thickness of the shock is very small the Mach number upstream of the shock may be taken to be 2.8 itself, okay. We are not going to account for the thickness of the shock. So, Mx may be taken to be 2.8 and from the normal shock table for Mx equal to 2.8 we can get My equal to this and the ratio of static pressure is to be 8.98. So, the exit pressure in this case works out to be 8.98 times 36.85 which is 330.91 kPa, okay. So, this solution let us use a slightly different color for this, okay. See, we have already used green maybe brown. So, this solution has an exit pressure of 330.91 kPa, okay. So, now we have the three pressure values corresponding to the three regimes. So, now we are in a position to proceed with the analysis. First case, since the given back pressure is 20 kPa and that is less than the design value. Notice that 20 kPa is even less than this value. So, 20 kPa would come somewhere there. So, that is even less than this which means that the flow is under expanded. There is some more expansion that needs to take place outside the nozzle and the values for the exit properties are the same as design values. There is no change in the flow field inside the nozzle. Part B, the exit, I am sorry, the back pressure is given to be 500 kPa which lies between as you can see which lies between this solution 330.91 and this solution which is 980.4. So, that means 500 comes somewhere over there. So, that means the solution that we are looking for is going to look something like this. So, there is a normal shock somewhere in the divergent portion of the nozzle. So, there is a normal shock in the divergent portion of the nozzle. There are couple of ways in which we can determine the location of the normal shock. That is most important and the exit mark number. So, we need to determine both these quantities. So, there are couple of ways in which we can do this. Let us go through the first one which is very accurate. Let us see. So, the mass flow rate at the throat section and the exit section may be written like this and they are equal. So, we rewrite density in terms of I mean using the equation of state and we do the same for the exit density also. And then if you rearrange this, you get something like this where you have taken the P to the left hand side. And if you simplify this expression further, the left hand side contains quantities which are all known. So, Ae over A throat is given to be 3.5 square root of gamma plus 1 over 2 known. P01 is 1 MPa or 1000 kilopascal. P exit is given to be because it is subsonic, the exit pressure is equal to the ambient pressure. So, the exit pressure in this case is 500 kilopascal. So, if you substitute the values, we get this expression from which we can solve for the exit mark number which comes out to be 0.327. And the exit stagnation pressure comes out to be 538.435 kilopascal. So, the stagnation pressure has decreased from 1000 kilopascal to 538.435 kilopascal. So, that is almost 50% loss of stagnation pressure. Now, if you actually look at the flow situation that we have and that we have drawn. So, let us draw it again here. So, normal shock sits somewhere over here. So, we label this as x and this state as y. So, P0 upstream up to I mean P0 upstream of the shockwave is 1000 kPa and P0 downstream of the shockwave is 538 kPa. So, P0 y over P0 x is equal to 0.538 435, 0.538 435. So, we may now go to the normal shock table. So, P02 over P01 0.538 45. So, that falls somewhere between these two. And so, the mark number upstream of the shockwave is between 2.4 and 2.5. So, by interpolation we can get the mark number to be 2.405. So, this is the mark number just ahead of the shockwave. Now, the flow is isentropic up to the shockwave. So, we can go to the isentropic table and retrieve A over A star corresponding to this value of mark number to be 2.414. So, the normal shock occurs at a location where A over A throat is 2.414. Remember, the flow is choked at the throat. So, A throat is equal to A star upstream of the shockwave. So, we now have a complete description of the flow field with based on the given ambient pressure which is 500 kPa. Now, an alternative method of solution for this is to actually start by assuming the location of the shockwave. So, basically what we do is the following. So, we guess the location of the shockwave. In this case, our first guess is that it occurs at a location A over A throat equal to 2. So, once we have this from the isentropic table, we can retrieve this value for Mx. This is the mark number just upstream of the shockwave. Once we have this mark number, we can then go to the normal shock table and calculate P0y over P0x where P0y is the stagnation pressure immediately downstream of the shockwave. Now, we may calculate A exit over A y star. Notice that when you have a shock, let us call this x and let us call this y as we are doing. Let me just draw this slightly differently so that we have more room here. So, when we have a normal shock sitting somewhere here and this is x and this is y. Notice that the sonic state upstream of the shockwave occurs at the throat. So, A x star is equal to A throat. Whereas, the sonic state downstream of the shockwave occurs at some other hypothetical location. Actually, the flow leaves the nozzle at a subsonic mark number. But if we want the sonic state, then we could actually do something like this. So, we could actually construct a conversion throat like this. So, this would be A y star. And of course, this is the exit. So, this is A e. So, A e over A y star may be evaluated like this. So, we write A e over A y star as A e over A throat times A throat over A y star. A e over A throat is equal to 3.5 that is given in the problem statement. Now, A throat over A y star has to be determined by using the fact that the mass flow rate, choke mass flow rate through this throat and the choke mass flow rate through the second throat are both equal. So, that is where this comes from. So, this is the part of the expression for the choke mass flow rate through the first throat. And we have cancelled out the expression involving the properties of air. And this is the mass flow rate through the second throat. This is the first throat. So, from this, you can get the fact that A y star over A throat is equal to P0 x over P0 y. Now, what is important about this is that the ratio P0 x over P0 y is greater than 1, which means that the second throat has to have a higher cross-sectional area to swallow the same mass flow rate because of the loss in stagnation pressure across the shockwave. It cannot be the same area as the first throat. It cannot be smaller than the area of the first throat. It has to be larger than the area of the first throat by an amount equal to or by an amount which is dependent on the loss of stagnation pressure so that it can swallow the same mass flow rate. It is a very important concept that comes from looking at this, I mean solving this problem in the alternative manner. So, I can now calculate, so I can now calculate A e over A y star with the value. So, I know P0 y over P0 x and A throat, I am sorry A e over A y star is also known. So, I have this value. Now, with this value, if I go to the isentropic table, I get the corresponding P0 e over P e and from which I can evaluate the exit static pressure. So, this has to be done in a sequence like this from left to right. So, we guess the area ratio or the location at which the normal shock occurs. Then we get Mx Mach number ahead of the shockwave, just ahead of the shockwave. Then you get the ratio of stagnation pressures across the shockwave from the normal shock table. Then we evaluate A e over A y star and then with that we go to the isentropic table to retrieve the value for P0 exit over P exit from which we can calculate P exit. Remember P0 exit itself is known. So, P0 exit is equal to P0 y and P0 y itself is known because the ratio P0 y over P0 x is known. So, that means P0 x equal to P0 1 which is given to be 1000 kPa. So, P0 e is just known. So, I can get the exit static pressure. Now, normally what we would do is, we will choose the initial guess for the shock location by making the following observation. Remember, so A e over A throat, now the actual location of the shockwave is between two places, here where the area ratio is 1 and here where the area ratio is 3.5. So, it lies between 1 and 3.5. So, typically we would take the arithmetic average of the two and then take that to be our first guess. So, that occurs right in the middle of the diverging portion. Now, we have not done that in this present example because that would have been too close to the final answer. So, I wanted two or three iterations to demonstrate how this procedure goes about. So, our next guess would be 2.5 for the area ratio. So, we go through the same procedure. So, corresponding to area ratio 2.5, we get the Mach number upstream of the shockwave to be 2.443. Then we go through the same procedure and we get our exit pressure to be 482. Remember, our target value is 500 kilopascal. So, we have now straddled. So, these two values straddle the target value of 500 kilopascal. So, we make a slight adjustment. We take area ratio to be 2.42 and we get the exit pressure to be 497 kilopascal which is probably close enough. The exact value that we obtained from the previous calculation is 2.414 whereas here we are getting 2.42. Further adjustments, of course, are possible but probably not worthwhile. This itself is acceptable. So, this procedure although it is somewhat more involved is more insightful because it allows us to understand for instance, you know the concept of the sonic state. The fact that the sonic state is different upstream of the normal shockwave and different downstream of the, I am sorry, different from what is there upstream of the shockwave to what is there downstream of the shockwave. And the fact that the second throat has to be larger than the first throat in order to swallow the same mass flow rate on account of the loss of stagnation pressure across the normal shockwave. So, this illustrates the importance of the sonic state as a reference state. Remember here the sonic state is this sonic state is a hypothetical sonic state whereas this one is real. This is happening at the throat of the nozzle. So, this demonstrates the power of the sonic state as a reference state. And also the importance of this expression here which is mass flow rate, choked mass flow rate through a nozzle.