 But the general trend is that the downstream Mach number M2 decreases with increasing M1 and as you can see here the static static temperature increases with increasing Mach number and this does not seem to be asymptoting to a constant value. It seems to just continuously and monotonically increase with M1, the same is true for P2 P2 over P1 also. So, the pressure after the shock continues to increase with Mach number. What is that? Although P2 is greater than P1, T2 is greater than T1, rho 2 over rho 1 which is equal to P2 over P1 times T1 over T2. So, this quantity is still greater than 1. Although this is greater than 1 and this is also less than 1, this quantity is still greater than 1 as you can see from here and this also seems to be trending towards both this and this seem to be trending towards an asymptotic value. The most important quantity of course is P02 over P01. So, you can see that there is an increase in the loss of stagnation pressure with an increase in M1 which is understandable. The shock becomes stronger and stronger as the initial Mach number increases. So, the loss of stagnation pressure is also quite high. So, you can see that for M1 equal to 5, actually for M1 equal to 5, the solution that we have derived is strictly speaking not valid because the temperature downstream of the shock wave would be very high and the gas would be far from being calorically perfect. So, this is an analytical solution, but sort of gives us an idea. So, you can see that at M1 equal to 5, the loss of stagnation pressure is almost 90 percent. But the interesting thing is for M1 equal to 2, the loss of stagnation pressure is only about 30 percent or so. It begins to fall steeply only after M1 equal to 2. So, in fact by allowing M2 to tend to infinity, M1 to tend to infinity, we can actually show that the asymptotic value for M2 is this which is what we are seeing here. And the asymptotic value for rho 2 over rho 1 is this which is what we are seeing here. And other values tend to infinity as M1 goes to infinity. Now, we turn to the issue of effectiveness or efficiency. So, in a normal shock wave, P02 is less than P01, obviously there is a loss of stagnation pressure. But the important thing is for isentropic compression process P02 is equal to P01 and that is highly desirable, no doubt about it. Loss of stagnation pressure implies irreversibility which implies loss of work. However, if you look at say a normal shock solution like this. So, for a given value of V1, specific volume V1 and V2, the pressure that is reached by normal shock compression process is over here. Whereas, the pressure as a result of an isentropic compression process would have been over here. What is that? This is s equal to constant. So, that is the isentropic line. So, isentropic compression process would follow this line and for a given V1 and V2, the pressure that we would reach is only this. Whereas, the pressure that we reach with the normal shock process is much higher. So, in a practical application like for example, the diffuser of a supersonic of the engine in a supersonic aircraft, given V1 and V2 or the specific volume that we are talking about corresponds to the size of the or length of the diffuser. Which means that for a given length of the diffuser, normal shock compression can accomplish a higher exit pressure when compared to isentropic compression although there is a loss of stagnation pressure. So, that is an important aspect. So, what will in the case of supersonic aircraft, weight is a very very critical performance criterion. The longer the intake, the more its weight and the more the drag will be. So, it should be short and we also would like the pressure at the end of the diffuser to be as high as possible and minimal loss of stagnation pressure. So, these are conflicting requirements. So, what I am trying to show here is that normal shock is very effective as a compression process. So, for a given change in specific volume, we can achieve a higher pressure or let us look at it the other way around. For a given pressure change, what is the we look at it this way. I am sorry. So, for a given pressure rise, so let us say for a given pressure rise which means P12 P12 P2. So, for a given pressure rise, the normal shock compression process is able to achieve achieve it with a certain change in specific volume whereas, the isentropic compression process. So, we go along the isentropic compression line until we hit this pressure P2. So, you can see that for the isentropic compression process, the change in specific volume is much more than the change in specific volume for the normal shock compression process which means that if you want to compress the flow isentropically, so that there is no loss of stagnation pressure, you require a longer diffuser. But longer diffuser as I said means more weight, more drag. So, although you may have saved, I mean you have avoided loss of stagnation pressure, you are actually paying the penalty somewhere else. So, normal shock compression in this case as you can see from both these cases, normal shock compression is very effective. But it is not as efficient as the isentropic compression process because of the loss of stagnation pressure. Now, if you look at this, I am sorry, if you look at this diagram here, you can see that the loss of stagnation pressure in a normal shock compression process is not actually bad until about m equal to 2. So, if you want to design a supersonic diffuser which can diffuse a supersonic flow from higher Mach numbers to let us say transonic Mach number, the best thing would be to accomplish that compression using using a mechanism other than normal shock until you hit a Mach number of 2. And once you hit a Mach number of 2, probably you can use a normal shock to achieve the rest of the compression because normal shock is actually not only quite efficient, but all I am sorry, quite effective, but also reasonably efficient for Mach numbers less than 2. So, that is the best compromise that we can have when you are actually designing the diffuser. So, typically what would be done is for Mach numbers above 2, the compression would be achieved, compression and deceleration of the flow would be achieved using oblique shock waves. And till you hit Mach number of 2, at which point you will terminate the compression and deceleration process using a normal shock wave. That is what is typically done in these supersonic diffusers. So, as I said, normal shock compression is more effective, but less efficient than isentropic compression because of losses stagnation pressure. Effectiveness is very important in practical design because it determines the physical dimension. The latter attribute which is efficiency, losses stagnation pressure is also important. So, you have to determine an optimal design like what I described just now. Let us work out an example involving a normal shock wave. So, air at 100 kPa and 300 Kelvin and moving at 696 meter per second encounters a stationary normal shock, determine the static and stagnation properties ahead of and behind the shock wave. So, static temperature and static pressure are given, the speed of the fluid. So, Pt, P1, T1 and V1 are all given. So, this is V1, this is T1, P1. So, P1 is given, T1 is given, V1 is given and we can evaluate the speed of sound A1 equal to square root of gamma RT1 that comes out to be 348. So, M1 comes out to be 2. Now, we can of course substitute this into the expression for T0 and P0 and evaluate or we can also use tabulated values for this. Let us see the gas table. So, here as you can see T0 over T, T0 over static temperature T, T0 over T and T0 over P are all given as functions of Mach number. So, we can easily go to this table and for instance retrieve this entry from the table for M equal to 2, we have T0 over T equal to this number and P0 over P equal to this number. So, we can now evaluate T0 and P0. Now, once I have M1, I can always do the following. Once I have M1, once I have M1, I can evaluate M2 from this expression. Once I have M2, I can then evaluate P2 over P1, T2 over T1 from this expression. But this actually can be made and done much easily by going to the normal shock table. So, the normal shock table lists for various values of M1, it lists M2, P2 over P1, T2 over P1, P02 over P01. For M1 equal to 2, I can go here retrieve M2, which is 0.577, P2 over P1, pressure rise factor is 4.5, T2 over T1 is equal to this. So, P2 over P1 is equal to 4.5. So, I can get P2, I can get T2 and P02 over P01 is 0.72. So, the loss of stagnation pressure as you can see is roughly about 38 or so, P02 over P01 is 0.7209. So, the loss is roughly about 30 percent or so, and M2 is also known. So, once M2 is known, the velocity of the fluid downstream of the shock wave may be evaluated. Notice that the velocity was 696 meter per second, it has now come down to 260.8 meter per second. So, the flow has been decelerated across practically zero distance from 696 to 260.8. So, that is the power of the shock wave. So, you can see these values are entered here. So, you can see that since the width of the shock wave is negligibly small, you can see how effective shock wave compression processes. So, if we wanted to decelerate a fluid like this, we would have used diffuser for example. So, diffuser would have a finite length, it will require a finite length to decelerate the fluid from 696 to 260.8 meter per second. Whereas, the shock wave does it with zero distance or does it in zero distance and accomplishes a compression increase in static pressure from 100 to 450 kilo Pascal nearly four and a half times. So, the purpose of the diffuser is to decelerate the flow and convert the momentum of the fluid, incoming fluid to pressure rise. So, if you look at how to do that effectively, the shock wave is very, very effective. It accomplishes this sort of deceleration and this sort of pressure rise across zero distance. Whereas, the diffuser requires a finite length to do that. When you have a finite length, that means you add to the weight of the engine and you increase the drag because as the fluid flows through the through a finite length diffuser, there is wall skin friction drag. So, now you can see how normal shock compression can be attractive for certain applications, even though there is a loss of stagnation pressure and in this case, we said loss of stagnation pressure is about 30 percent or so, it can still be attractive because it is extremely effective. So, this completes our discussion of one-dimensional flows. What we will do in the next lecture is start our discussion of quasi one-dimensional flows, which is essentially flow through nozzles.