 So, what we will do is we will continue with what we left off in this compressible flow discussion yesterday, just a quick recap of what we were doing yesterday. We derived the speed of sound relation and also we have gone through the analysis of the normal shock phenomenon. And at the end of the normal shock discussion, we have introduced these normal shock tables, which are also very useful and convenient way to solve problems which involve normal shock situation. So, with this background now what we will do is we will move on and talk about one of the most important parts of introduction to compressible flow and that is a steady quasi one-dimensional isentropic flow and that is what I have projected on my slide right now. So, quasi one-D is supposed to mean almost one-D which is basically to convey the idea that we are talking about a flow inside a duct whose cross-sectional area is varying really slowly as you move along the flow direction. So, what I have shown here is a section of the duct and this is what I am calling a control volume. You can think about this as a semi-infinitesimal control volume. Semi-infinitesimal is because along the flow direction we are going in an infinitesimal distance dx, but across the cross-section we are spanning the entire cross-section of the duct. So, the duct is changing its cross-sectional area from A to A plus dA over a distance of dx and that is what I mean by slowly varying cross-sectional area. So, this dA is infinitesimally small and in that sense what we can assume is that as the flow proceeds from one axial location to the next axial location the lateral component of the velocity is going to be very, very small and predominantly the flow is characterized by a component of velocity in the axial direction and these arrows are supposed to show the axial direction. So, what we have is a control volume whose left face has an area A and the right face has an area A plus dA. What we have is a flow coming into the control volume at the speed of V and with a Mach number m and the conditions in terms of the flow variables are P, T and rho and through the distance dx all these changes by infinitesimally small amount dP, dT, etcetera and this goes on from the entrance length to the exit length. We talk about this situation as an adiabatic and frictionless flow. So, let me just quickly mention what we mean by frictionless flow. We are talking about the fluid friction or in other words we are basically saying that we are going to ignore the presence of viscosity altogether and we are treating this as an inviscid flow. It is a fairly useful model to employ although as all of you know in real life viscosity is always going to be present, but if you are talking about a duct which is relatively short in length the viscous effects can be reasonably well neglected and under that assumption we are going to perform our analysis. So, that is why it is adiabatic and frictionless flow because the fluid friction through the absence of viscosity is not present we will essentially ignore all the irreversibilities which are associated with frictional effects and therefore, this is going to be treated as an isentropic flow situation. So, to analyze the situation what we will do is with respect to the control volume that I had shown on the previous slide we will employ our mass balance and momentum balance as what was done yesterday. The only thing that one has to remember here is that the area of cross section is changing from A to A plus D A and accordingly you have to take into account the mass balance and the momentum balance. So, going to my next slide what I have written here on the left hand side where my cursor is sitting right now rho times A times V is simply the mass flow rate that is coming into over control volume and the right hand side simply signifies the mass flow rate that is going out of the control volume. What is normally done is you expand this right hand side and again neglect the higher order term and eventually you divide the entire equation by the product rho times A times V and what you will end up with is what we call a differential form of the mass conservation equation as applicable for this steady quasi one dimensional isentropic flow and that is what is written in the box here where my cursor is standing. When it comes to the momentum balance which is on the right hand side all that we realize and write is that the net force acting on my control volume is going to be equal to again the momentum rate of momentum leaving minus the rate of momentum coming in and that is how this m dot times V plus dV minus m dot times V is going to signify these are the rates of momentum leaving minus the rate of momentum coming in which can be immediately simplified noting that m dot is equal to rho times A times V and that is what you are going to see at the right most part of this equation. You can calculate the axial force net axial force acting on my control volume let me go back to this for a minute. So, what we have is pressure acting on the left face in the positive x direction let us say pressure P plus dP acting on the entire area A plus dA in the negative x direction on the right face and also in this particular case you have to include the axial component of the lateral force that is acting on this side area or the lateral area which my cursor is highlighting right now. If you work out the algebra you will see that the lateral force whose component in the axial direction is what we are interested in comes out to be simply P times the change in the area that is P times dA and the remaining forces are what are written on the left hand side. You go ahead and simplify this equation and the form of the momentum balance equation applicable for our steady quasi one-dimensional isentropic flow turns out to be this dP plus rho times V times dV equal to 0 and immediately for the purpose of our analysis I am going to express this momentum balance in a particular fashion. So, what is done there now is I write this as V times dV equal to minus dP over rho and the left hand side V times dV is written as V squared over V whole thing multiplied by dV. So, it is the same thing and on the right hand side I introduce a dP over d rho times d rho over rho. So, it is exactly the same as what it was. Now, given that we are working already within the isentropic flow situation this dP over d rho is by definition or by design going to be at constant entropy and therefore, from our previous discussion it is simply square of the speed of sound. So, I am replacing this dP d rho with a square and then when you take the ratio of V squared to a square you will get a square of the Mach number. So, the final form of the momentum balance equation which I am going to utilize to complete my analysis is boxed at the right bottom and eventually what is done is you eliminate this d rho over rho which is present between the continuity or the mass balance and the momentum balance and finally, what you will get is this so called area velocity relation which will tell us how the velocity of the fluid is going to behave with respect to the change in the area of cross section depending on what Mach number of the flow exists. So, this relation which is boxed at the left bottom is a very important relation and many of you would probably remember seeing it as well we call this the area velocity relation. So, let us see how we can interpret this area velocity relation remember that the velocity at any particular cross section which is given by a here is going to be positive. So, V as well as a are always going to be positive the Mach number can be subsonic which will which means that it is going to be less than 1 it can be sonic which is exactly equal to 1 and it can be supersonic as well which means it can be greater than 1. So, now let us see how we can interpret this situation physically let us say that we are dealing with a subsonic flow situation which means that the Mach number is less than 1. In other words the denominator m squared minus 1 is going to be negative. So, in this case what we will see is that if the area of the cross section is decreasing which means that d a is negative the negative over negative will provide a positive value or in other words what we will see is that if we are dealing with subsonic flow and if the area of the cross section decreases this will imply that the velocity of the flow is going to increase that is the first line here. On the other hand if you are dealing with a supersonic flow m squared minus 1 is going to be positive and therefore, in this case if the area of cross section decreases which means that d a is negative what we actually see is that the flow velocity is also going to decrease. So, this is something that we are more or less familiar with when it comes to the situation of Mach number equal to 1 what we will realize is that the denominator here will actually become 0 and in order to have any finite limit or finite physical meaning out of this expression we will say that unless and until d a is also going to be 0 we cannot really have a finite limit from the right hand side. In fact we force this condition and say that if m equal to 1 if the Mach number is going to be sonic the left hand side d v over v is going to be finite only with the condition that the area is minimum in some sense minimum or maximum whether the d a is going to be considered in that sense, but one way or the other we say that unless d a over a is going to be 0 we are not going to have a finite limit for d v over v if the Mach number is going to 1. So, now putting these three points together what we realize is that if we want to accelerate a flow from a subsonic velocity to a supersonic velocity in this quasi one dimensional isentropic situation what we must have is initially a convergent duct resulting finally in a minimum area location where the Mach number is going to become 1 followed by an area increase or a duct increasing in cross sectional area because that will further increase the Mach number from 1 to higher supersonic value and that is what is written in my fourth bullet here on the slide that to accelerate the flow from a subsonic to supersonic velocity we must have a convergent divergent duct with an area minimum which is by the way denoted by a star and we will come to the star terminology little later, but this area minimum is what is called as the throat as many of you would know this terminology where the Mach number of 1 is going to be occurring. So, the only one point that we must keep in mind here is that just because 1 is going to have a throat in your convergent divergent duct that does not necessarily mean that automatically m or the Mach number is going to be 1 there. So, the Mach number is going to be 1 depending on the overall pressure difference that is driving the flow if the pressure difference is not sufficient you may not get m equal to 1 at the throat. However, physically we have an area minimum at the throat and therefore, using the area velocity relation itself again what we can definitely conclude is that if m equal to 1 is not going to be reached anywhere if we have totally subsonic flow we will have the maximum velocity at the throat and if we have totally supersonic flow we will have the minimum velocity at the throat. However, if Mach number of 1 has to be reached in this convergent divergent situation it must be reached at the throat that is the way to interpret this relation. What we have done so far is we have employed again going back to my previous slide a mass balance and our momentum balance for the control volume that I had drawn. If I go ahead and employ the steady flow energy equation noting again that there is no heat and work transfer terms and neglecting any changes in the elevation what we have is h plus v squared over 2 remaining a constant. So, h plus v squared over 2 coming in is the same as h plus v squared over 2 that is going out of the control volume. If you end up simplifying this equation by discarding the higher order term what you will be left with is h plus v times dv equal to 0 and this v times dv is something that you can write as d or the differential of v squared over 2 and therefore, you can actually integrate this equation and say that h plus v squared over 2 must be a constant in this steady quasi one dimensional isentropic flow and that constant is by definition denoted as the stagnation enthalpy. So, the stagnation enthalpy h naught by definition is equal to h plus v squared over 2 which is a constant in our steady quasi one dimensional isentropic flow. So, formally what do these stagnation conditions mean? Stagnation conditions are supposed to mean that the following that you consider a situation like our quasi one dimensional isentropic flow at any particular cross section you have certain conditions of the flow given by the Mach number, the velocity, the pressure, the temperature and so on. And now you imagine that from those conditions the flow is brought to rest or other words to zero velocity in an isentropic manner. So, at the end of this process where you bring the flow to rest whatever conditions you achieve for pressure, temperature, density, enthalpy, etcetera will all be called as the stagnation value. And remember that we are bringing the flow to rest which means that the velocity is going to be equal to exactly zero at the end of this slowing down process. So, therefore, going back to my stagnation enthalpy expression if you just evaluate this h plus v squared over 2 at the stagnation condition since v is going to be exactly equal to zero what you are left with is the stagnation enthalpy itself. So, that is the way I would like to interpret this h plus v squared over 2 as the stagnation enthalpy. Now, there are a few more important and useful results as far as problem solving is concerned those you can immediately bring about from these stagnation conditions and one such derivation is outlined at the bottom of the slide. So, since we are dealing with ideal gases we are going to express the enthalpy as C p times the stagnation temperature. Likewise, the stagnation enthalpy will be expressed as C p times the stagnation temperature. So, stagnation temperature is again exactly the same namely again going back to the definition of stagnation condition at some location within your convergent divergent duct let us say where steady quasi one dimensional isentropic flow is happening. The temperature at some location is T imagine that the flow from that location is brought to rest in an isentropic manner. So, whatever the temperature value is achieved at the end of the slowing down process to rest we are going to call that as the stagnation temperature. So, going back to the slide then again stagnation enthalpy simply expressed as C p times the stagnation temperature and by suitably manipulating this equation what you can obtain is the ratio of the stagnation temperature to the so called static temperature in terms of the Mach number that existed and the gamma. So, what this supposed to mean is that at again at some particular cross section that we are focusing on within our convergent divergent duct let us say that the conditions given by the temperature are T and the Mach number is m. Knowing these two values the Mach number and the temperature at that particular location you are in a position to figure out quickly what is going to be the value of the stagnation temperature using the expression which was just derived and is sitting at the bottom left. Now, this is a ratio which is very important T naught over T. Since we are talking about isentropic situation I can relate T naught over T to P naught over P using the standard isentropic relations and that is simply you raise the T naught over T to gamma over gamma minus 1 and therefore again looking at how the expressions are written if I know at a given location the pressure and the Mach number I am in a position to evaluate what is going to be the stagnation value of the pressure in this quasi one dimensional steady isentropic flow using the second boxed expression. So, these ratios T naught over T P naught over P likewise you will have rho naught over rho the stagnation density divided by the density itself as a function of Mach number and the gamma are all very important when it comes to problem solving. So, going ahead there is one more reference condition which normally we talk about is and that is what is called as a critical or sonic condition. So, the stagnation condition was that from a given flow situation we imagine the flow to be brought to rest in an isentropic manner. Instead of bringing it to rest in case of these critical or sonic conditions what we imagine is that we start the flow from whatever condition it is in and bring it in an isentropic manner to a condition of Mach number of 1 and that condition when the Mach number of 1 is achieved is what is called as the sonic or critical condition and just like the stagnation condition was denoted by a subscript 0 as in the previous slide the critical or sonic condition is denoted usually by a superscript star. So, now let us go ahead and utilize the previous expression that we had obtained as the ratio of the stagnation temperature to the temperature in terms of the Mach number and gamma and what I am going to do now is let us put the condition that the Mach number at the location where we are looking at is itself 1 corresponding to that Mach number of 1 we must have the temperature equal to the critical temperature or the sonic temperature you can call it that way. In other words if m is equal to 1 t is going to be equal to t star and by simply substituting m equal to 1 in this expression I am in a position to calculate this one other important ratio t naught over t star and it simply turns out to be gamma plus 1 over 2 in this particular situation. Once we know t naught over t star since we are still dealing with isentropic flow using the isentropic expression I can find out what is the ratio of t naught over p star the stagnation value of the pressure to the critical value of the pressure and that is simply going to be gamma plus 1 over 2 raise to gamma over gamma minus 1. So, usually you will see that in books if we are specializing only to a gamma equal to 1.4 situation which is the gamma for air you can immediately evaluate these quantities. So, this will be 1.4 plus 1 that is 2.4 over 2 and this will be then 1.4 plus 1 over 2 raise to 1.4 over 0.4 etcetera and these can be figured out as constants depending on what value of gamma you are using most of the times we restrict ourselves to a use of gamma equal to 1.4 and in that sense the p naught over t star and p naught over p star can be immediately figured out as the constant values corresponding to gamma equal to 1.4. I have not bothered to put those constants here, but it is pretty obvious that if you substitute gamma equal to 1.4 here you will come out with a constant value. Finally, a very important quantity to evaluate in these internal flow situations is the mass flow rate and if you look at any particular cross section within our convergent divergent duct the mass flow rate is simply going to be the product of the density at that location, the velocity at that location and the area of cross section at that location within our cos i one dimensional approximation. So, I want to express it in a particular fashion. So, I write this as a mass flux or a mass flow rate per unit area which is then simply the product of density times the velocity. Density I express using the ideal gas law and I introduce these gammas in an appropriate fashion to finally, express the mass flux or the m dot over a in terms of the pressure, the temperature and the mark number at the location that we are looking at. And finally, since I already know the relations between the pressure and the stagnation pressure also temperature and the stagnation temperature going back here. So, these are the relations I know that p naught and t are related in this fashion p naught and p are related in this fashion. So, substituting for p and t in terms of p naught and t naught and the mark number if you go ahead and simplify the relation what you will end up obtaining is that m dot over a is going to be equal to all these quantities on the left and the right hand side which will basically involve your stagnation pressure and the stagnation temperature. So, we can see here is that let us say the mark number is something that is a constant value for the purpose of our discussion. Then you will see that if you increase the stagnation pressure somehow or decrease the stagnation temperature you are in a position to increase the mass flow rate in general. So, this is a very important relation at the bottom of your slide. This is what I will call as a mass flux expression m dot over the area of cross section which is written in terms of gammas and r's and the mark number which is existing at that cross section where we are calculating this. And the two constants of this problem which are the stagnation pressure and the stagnation temperature. So, this is an important relation and for many a problem you will utilize this mass flow rate expression for evaluating it. So, now let us see how we can utilize this further. Noting that p naught and t naught are going to be constants for our steady quasi one dimensional isentropic flow situation. What I want to do is I want to relate the area of cross section say in section 1 corresponding to which the mark number will be m 1 and the area of cross section a 2. Let us say in section 2 corresponding to which the mark number is going to be m 2. Because it is the same mass flow rate that is flowing between section 1 and 2, this m dot is a constant and using that m dot equal to constant and employing this mass flux relation between two sections 1 and 2 in your isentropic duct. So, here is section number 1, here is section number 2. Let us say m 1 here, m 2 here, a 1 here, a 2 here, m dot being the same you just take the ratio in appropriate fashion to come up with a 1 over a 2 equal to a function of m 2 over m 1 or in other words this is just going to be a function of the two mark numbers at the two cross sections and the gamma. And now let us say that one of these two in particular this a 2, I will force to be a cross section area where the mark number is equal to 1. So, let us simply assume that in the cross section 2 we say that mark number 1 is achieved. In other words cross section 2 is the critical cross section or you can say a throat section. So, using that I drop the subscript 1 on the numerator and relate any general cross section a with the critical cross section a star in the form of this ratio a over a star. So, all that I am doing is instead of a 2 I am calling this as the critical cross section corresponding to which m 2 is going to be exactly equal to 1 because a 2 is equal to the critical cross section m 2 is going to be equal to 1 there and there if you simplify the relation you get a over a star as a function of m and gamma. So, let me go back to my board I did not draw the function here on the slide let me try to draw it here. So, roughly if you plot that function it will come out to be something like this value will be equal to m equal to 1 this will be equal to 1. So, where m is equal to 1 the a over a star is equal to 1 or in other words we are dealing with the critical cross section itself. On the other hand any other value that you want to think of for the area. So, let me draw a horizontal line in general you will see that this intersects the curve in two different locations or in other words if I have a value of area cross section which is other than the critical cross section this corresponds to two different values of the Mach number. So, going back to my board this first Mach number is subsonic and the other Mach number is supersonic. So, this is something that we will utilize quite a bit that you can imagine a convergent divergent duct. So, that at two sections the a over a star value can be identical but one will correspond to a subsonic value of the Mach number whereas the other one will correspond to a supersonic value of the Mach number. So, in that sense this a over a star relation as a function of m is very important again to summarize any value of a other than a star you are going to have two possible isentropic flow solutions one of the two solutions is going to be subsonic Mach number the other solution is going to be corresponding to a supersonic Mach number. So, going back to my slide again that is precisely what I have summarize here what I just what I just said namely that for a given a over a star value not equal to 1 two values of Mach number are permissible within the isentropic theory one is going to be a subsonic solution and the other one is going to be a supersonic solution. Finally, just like what you had in case of the normal shock tables these expressions which we have more or less derived on our way are usually tabulated in what are called as the isentropic flow tables and again these are available for a given a gamma value. I have uploaded on Moodle a few days back the isentropic tables corresponding to gamma equal to 1.4 and what we have is again going back to my expressions here. For example, if you see p naught over p it is a function of M and gamma. So, let us say we fix our gamma equal to 1.4 and for a variety of values of M starting right from 0 all the way to wherever you want you can come up with the expression or the value I should say of the ratio p naught over p. Likewise you can come up with a value of the ratio t naught over t also you can do the exact the same thing for rho naught over rho and so on and these ratios are then tabulated in the isentropic flow table. So, the left most column that you will see is the Mach number and the value of the Mach number here will go from right from 0 remember that for normal shock table the Mach number will begin only at 1 because a shock situation is possible only when the supersonic or the upstream flow is supersonic with respect to Mach with respect to the shock I am sorry. Here on the other hand there is no such restriction the Mach number can start right at 0 and for each of these M values starting from 0 you have tabulated in the isentropic flow table p over p naught rho over rho naught a t over t naught and also this a over a naught. So, again remember that for this a over a naught if you choose one value you are going to have two possible Mach number one is going to be in the subsonic range and other one is going to be in the supersonic range. In fact all these ideas become really really clear when you want to solve the problems that have been provided in the exercise sheet. Problems number Cf 8 to Cf 15 will essentially deal with not only the isentropic flow problem discussion that we are going through, but also combining this with the normal shock discussion that we had and some of them the problems that is are not very obvious and you will have to spend reasonable amount of time to make sure that you understand how to solve it. But using these two tables the normal shock table that was provided yesterday and the isentropic table that we are talking about right now and which have already been uploaded on model you should be able to solve all the problems provided in the exercise sheet. Now the isentropic flow tables which we have uploaded on the model have a few more quantities also which I have not listed on the slide here and you can ignore those other quantities those do not pertain to the quasi one dimensional isentropic flow situation. As far as the quasi one dimensional isentropic 1D situation as what we are discussing here is concerned whatever I have listed on the slide here is essentially what you would like to utilize from that isentropic table. So, with this background let us just quickly go through the operating characteristics of these somewhat important devices called the nozzles. As many of you would already know what the nozzles do is that they act as propulsion devices and here there is an interchange between the enthalpy and the kinetic energy. Because as you can see that the energy equation provides that the enthalpy plus v squared over 2 which is simply the specific kinetic energy is a constant. So, what we end up doing is that the flow is expanded through these nozzles such that the enthalpy reduces and it is the reduction in the enthalpy is compensated by increase in the kinetic energy. So, that the flow accelerates from a low value at the beginning to a high value when it exits the nozzle and using that high value jet which exits the nozzle you can generate a thrust that will eventually propel the nozzle. So, you can attach this propulsion device to something like an airplane as we can imagine and then provide the thrust necessary to propel the aircraft. Obviously, what we are going to do here is a fairly basic discussion in terms of what happens when there is a certain upstream pressure and a certain downstream pressure using which the nozzle is operating and in particular if you keep the upstream pressure a constant and keep on reducing the downstream pressure how the nozzle properties in terms of its operation change is what we will want to discuss. So, what I am going to do is I will discuss the situation for a purely convergent nozzle first and then I will go on and discuss the situation for the convergent divergent nozzle. So, in fact, I am going to utilize mostly my board here to draw the situation. So, going back to my board let me draw a purely convergent nozzle in essentially an axis symmetric situation. So, this is my axis of symmetry and therefore I am not drawing the bottom part. So, what I have is a purely convergent nozzle in the sense that the area of the cross section of the nozzle from the inlet to the outlet is monotonically decreasing and will reach a minimum value at the exit. So, let me draw what we will call an inlet plane and an exit plane. Here I have inlet plane and here is I have exit plane. So, usually the way these devices are described is that the inlet plane is supposed to be attached to a large reservoir which has values of p naught and t naught the pressure and temperature equal to the stagnation value. So, the conditions that will be existing where the inlet plane is attached to a large reservoir will be equal to p naught, t naught, rho naught, etcetera and the velocity will be essentially equal to 0. So, what we are essentially assuming that this device the purely convergent nozzle right now is attached to a very large reservoir which contains pressure at p naught, temperature at t naught, density at rho naught and the reason why we call these as stagnation properties is because the velocity will be assumed to be basically equal to 0 within that reservoir. So, there is a very standard usage when it comes to these nozzle operation that we will call these stagnation conditions equivalently as reservoir conditions. So, let me write this that the stagnation condition is equivalently called as reservoir condition. So, when it comes to problems if a problem is mentioning that the reservoir conditions are so and so we are supposed to understand that those are nothing but the stagnation conditions. So, what we have is the inlet plane where the flow gets into this nozzle is attached to a large reservoir where the conditions are given by the stagnation value and the velocity is 0. The velocity then gradually is increasing through the nozzle and will reach a maximum value in this case at the exit plane. So, this is my exit plane and the velocity which will be exiting the nozzle is something that I will call as v suffix e corresponding to that there will be a mark number which will be m suffix e and the exit plane pressure which is going to be p e likewise the exit plane temperature can be p e. Finally, the setting is completed using a so called back pressure. The back pressure is essentially the pressure of the reservoir into which the nozzle is discharging. So, usually you can think about this as a large atmospheric reservoir which is maintained let us say at some atmospheric value if at all and then to that is the nozzle discharging. So, this is what we will call a back pressure which is the pressure maintained in the large reservoir into which the nozzle is discharging and usually the operation of such nozzles is described in terms of keeping these stagnation values constant and reducing this back pressure. So, what will happen is that if this back pressure is equal to the stagnation pressure there is absolutely no pressure difference across the nozzle because p naught will be equal to p b in that case and therefore, there cannot be any flow because there is no driving potential for the flow. Then you start gradually reducing the back pressure using that you are generating more and more driving potential for the flow and as the difference p naught minus p b increases you will basically start flowing more and more flow through the nozzle. So, let me go to the next page and try to describe this situation using a very standard plot where we express the pressure that is acting throughout this nozzle at different cross sections along the length of the nozzle as we change the back pressure value from p naught to lower and lower values. So, usually this is what you will see. So, this is my exit plane again and this is the back pressure on the back side let us say. So, this is the distance along the nozzle let us say and x equal to 0 is where my inlet plane is and usually I will show p over p naught equal to let us say 1 and from this value you will see how things are going to change. So, as I said the first situation is if you maintain the back pressure equal to p naught there is absolutely no flow and the reason is because there is no driving force or the driving potential in terms of a pressure difference that is available for the nozzle to function. Later on let us say you start reducing the back pressure. So, as you decrease the back pressure like I said earlier more and more driving potential will be provided and the flow will initiate and what will be seen is that typically the pressure profile through the nozzle will look something like this. So, you are continuously reducing the pressure from the inlet to the outlet and likewise you are also going to reduce the temperature. So, this is what we call an expansion through the nozzle. The velocity as the pressure and the temperature are decreased through the nozzle is going to continuously increase and through the area velocity relation that we discussed because this is an area minimum at the exit plane you will reach the maximum value of the velocity at the exit plane. And again remember depending on how much is the value of this p naught minus p b whether the flow in the exit plane is going to be reaching mach number of 1 or not will be decided. It will always be a maximum velocity situation in the exit plane that is perfectly fine. However, depending on the driving force or the driving potential p naught minus p b whether the mach number is going to reach its value of 1 or not in the exit plane is what we will look at. So, as you keep on reducing the back pressure the mass flow rate through the nozzle will keep on increasing and the pressure through the nozzle will keep on decreasing likewise the temperature will keep on decreasing. So, this will happen until you reach a situation which is corresponding to the value p star or the critical pressure corresponding to your stagnation pressure. So, if you reduce the back pressure all the way to p star which corresponds to your stagnation pressure then because we are dealing with an area minimum at the exit and because the pressure is now reached to value of p star we will reach a mach number of 1. So, here when p b is equal to p star the exit mach number will eventually equal to 1. So, let me draw one more line at the top. So, this is at yet lower value of back pressure. So, this way you decrease the back pressure and increase the mass flow rate and also increase the mach number continuously. So, finally, when the back pressure is reduced to such an extent that the exit plane pressure is equal to p star then we will say that the mach number at the exit is going to be equal to 1. At this point what happens is that the mass flow rate corresponding to this curve will be written as m dot star let us say. Why m dot star? Because we have reached the mach number of 1 at the exit. So, using that mass flow rate expression earlier if you knew the exit area you can actually calculate it using m of equal to 1 at the exit. Now a typical phenomenon that you will remember when this mach number of 1 is reached at the exit plane is that if you further decrease the back pressure below the value of p star at the exit plane. What we say is that no change occurs in the nozzle in the sense that the nozzle is choked or it remains frozen and the mass flow rate does not change from that m star m dot star value that was reached when the back pressure was equal to p star as we said. So, this is what we call a choking phenomenon. So, choking of nozzle this occurs when the back pressure is equal to p star. If the back pressure is below p star what we see is that the exit plane pressure remains frozen at p star the mass flow rate remains frozen at m dot star and this is what we call choking of the nozzle and corresponding to this choking we will also note that the mach number at the exit plane remains equal to 1. So, here is a situation where the back pressure which you were somehow reducing through some external mechanism let us say was getting continuously lowered it got lower to an extent that the exit plane pressure became equal to p star when the mach number at the exit plane became equal to 1 and the mass flow rate became m dot star and that is the maximum mass that you can pass through this nozzle mass flow rate that is. Further if you decrease that back pressure what we saw was the nozzle getting choked and the characteristics of this choking are written on my board right now that the exit plane pressure remains frozen at p star and m dot remains frozen at m dot star and the exit mach number remains frozen at m equal to 1. So, let me draw this situation on that plot and then we will try to explain why this choking occurs. So, this was the plot that I was using. So, let us say this was my p star and now my back pressure is something here I have maintained the back pressure which is below the value of this p star. So, nothing really changes inside the nozzle whatever were the conditions inside the nozzle when this back pressure was here remain exactly the same when the back pressure is now here. So, the pressure reduction from the value of this frozen value of p star at the exit plane to the lower value of the back pressure that you are maintaining outside the nozzle usually occurs through usually meaning always occurs I should say through what is called as expansion waves outside the nozzle and if the nozzle is working in such a situation that your back pressure is lower than this p star value which is frozen then this situation is what is called as an under expanded nozzle. The reason we use the words under expanded is because outside the nozzle. So, this is all outside the nozzle here outside the nozzle the pressure is able to further go down or the flow is able to further expand and therefore, the nozzle is considered to be under expanded. But remember that these expansion waves that occur outside the nozzle in this under expanded situation are inherently multi dimensional in nature and usually in this first introduction to compressible flow where we are describing everything within our quasi one dimensional framework you do not really analyze this expansion wave situation at all. So, usually what I will want to inform the students that this is how it happens and in case we want to know little bit more on the expansion wave we have to go to a compressible flow type course later where these phenomena are discussed in more detail. But as far as the present discussion is concerned what we need to know is that the back pressure is getting continuously decreased it reaches till the exit plane pressure becomes equal to p star corresponding to whatever p naught you have in the upstream reservoir. Then further if you keep on decreasing the back pressure the nozzle chokes in the sense that nothing changes inside the nozzle whatever change that is supposed to occur will occur only outside the nozzle through this expansion wave where the pressure will decrease from a p star value in the exit plane to whatever lower value of p b that you are maintaining outside the nozzle. Now this is what happens if you are operating the nozzle with p naught constant and reducing that back pressure continuously. Now the standard question that one wants to answer here is that what is the physical reason why the nozzle chokes and for that we cannot really explain that based on purely thermodynamic principles unfortunately we have to include a fluid mechanics knowledge into that and specifically it is the knowledge of the wave propagation that is required to answer that question. So let me try to answer it in a fairly brief manner if you remember what we were doing is that we were continuously decreasing this back pressure until the value it became equal to p star. Now further if you want to reduce the back pressure to a value below this p star what ends up happening is that essentially a disturbance is created at the exit plane where the exit plane pressure is equal to p star and the back pressure is now suddenly reduced little bit below that p star value when it was exactly equal to p star we were reaching m equal to 1 at the exit. Now we have reduced it below the value of p star this action of reducing the back pressure essentially creates a disturbance which tries to travel everywhere including up the nozzle in the form of an acoustic wave. Now what will happen is that the acoustic wave will try to move at the local acoustic velocity or the local speed of sound with respect to the flow speed but look what has happened when we reached m equal to 1 at the exit the velocity of the flow itself which is exiting this way becomes exactly equal to the sonic velocity whereas when you are trying to reduce this back pressure the acoustic disturbance that you are generating is trying to move up the nozzle with the same velocity namely the acoustic velocity. So what ends up happening is that the flow is moving out with the acoustic velocity because m is equal to 1 at the exit and this disturbance which is in the form of an acoustic disturbance or a sound disturbance is trying to move up the nozzle with exactly the same velocity equal to the sound velocity. So these two people just cannot go anywhere in the sense that the velocity of the flow is exactly equal to the sound speed. So against that flow velocity the sound wave that is generated as a disturbance wave when you reduce the back pressure is simply not able to move up the nozzle and unless and until this sound wave is able to move up the nozzle no condition inside the nozzle can change and this is the formal explanation why the nozzle chokes just to again recap the whole thing. We create an acoustic disturbance when we reduce the back pressure if the back pressure is already equal to p star if you further reduce it the acoustic wave which will try to move up the nozzle at the speed of sound simply cannot go anywhere because the exit plane flow velocity itself has reached a value of m equal to 1 and the speed of sound so that they are exactly equal and opposite to each other. Therefore against that incoming flow which is at the sonic velocity the acoustic disturbance which is also at the sonic velocity but trying to move in the opposite direction simply cannot go anywhere and change the conditions within the nozzle. So this is the formal explanation why a nozzle chokes as I said I have tried to explain it in as physical terms as we can little more details if you want you have to refer to a standard compressible text compressible flow textbook but as far as my opinion is concerned when you are dealing with compressible flow in a thermodynamic score somewhat whatever I just talked about is probably a sufficient explanation for the student. You cannot give all sorts of details in a thermodynamic score when it comes to specific fluid flow phenomena like the wave propagation and etcetera are concerned but at least to a good extent this provides some physical field as to what is happening. And the choking phenomenon whether it is a purely conversion nozzle or later now what we are going to discuss is a convergent divergent nozzle is exactly the same. There is absolutely no difference between the explanation that is provided why a purely convergent nozzle chokes and why a convergent divergent nozzle. So it is to do with the fact that an acoustic disturbance which is generated as a result of lowering of the back pressure etcetera is unable to move against a flow velocity which is also exactly equal to the sonic velocity and is in the opposite direction. So that is what ends up happening and therefore the nozzle chokes. So this is more or less what I wanted to point out in the purely conversion nozzle. So just to again recap the whole discussion what we have is a purely conversion nozzle. So the area of cross section monotonically decreases and becomes equal to the minimum at the exit plane. You have an upstream pressure which is equal to the stagnation pressure kept constant and the back pressure which is the pressure existing in the reservoir to which the nozzle is discharging is slowly reduced starting from a value of p0. So if pb is equal to p0 there is absolutely no flow as you start reducing the value of p0 below pb I am sorry below p0 the flow will initiate as you keep on reducing p0 the flow will keep on increasing in the mass flow rate correspondingly the velocity at the exit plane will always remain maximum but unless and until the pressure in the exit plane remains does not reach p star the Mach number at the exit plane will not reach m equal to 1. Only when the back pressure is reduced to such an extent that the exit plane pressure becomes equal to p star the exit plane Mach number will become 1 the mass flow rate then is at its maximum which is given by m dot star. If you further reduce the back pressure the nozzle is already choked in the sense that no change in the mass flow rate and the conditions within the nozzle can take place and we have now discussed the reason for it that the acoustic disturbance which will be generated when you reduce the back pressure is simply not able to move into the nozzle because the exit plane flow is exactly equal to the sonic velocity and in the opposite direction. Once the back pressure is reduced below the value of p star in the exit plane the nozzle operates in what we call as an under expanded mode where the expansion from p star in the exit plane to the lower value of p b that you are maintaining outside the nozzle will take place through what we call as expansion mode. So, this is how the purely convergent nozzle functions. If this was reasonably followed the discussion for the convergent divergent nozzle is not that troublesome other than the fact that there is one more additional feature which will show up in the discussion of the convergent divergent nozzle and that is the formation of a shock in the divergent nozzle for certain operating conditions.