 So, we will discuss normal shock waves in this lecture. Normal shock waves are compression waves, this is very important, normal shock waves are compression waves that are usually seen in nozzles, turbomissionary blade passages, supersonic intakes. So, these are intakes that are provided in front of the engine of supersonic aircraft like SR 71, Concorde, F-15 and so on, they are also seen in shock tubes. In fact, the reason for discussing shock waves in this course is precisely because they are encountered in nozzles. Since our focus is on learning or understanding compressible flow through nozzles and since this appears in nozzles regularly, we need to learn about normal shock waves. But normal shock waves are seen in external flows also, flow over aircraft, flow over aircraft wings and so on. Good example of that is shown here. I must apologize that I am not able to ascertain the source of this picture but you can see here the picture of an F-14 which is a supersonic aircraft as it breaches the sound barrier. So, you can see this vapor halo which is actually condensation of water droplets but that shows what happens when the aircraft starts to move at supersonic speeds. So, you can see these types of shock waves only when the aircraft begins to move with supersonic speeds or in a frame of reference where the aircraft is stationary, the air moves with supersonic speed over the stationary aircraft either ways. So, this is a very nice illustration of a shock wave and this is in external flow. The same sort of thing is seen in internal flows also as I just mentioned nozzles, turbo missionary blade passages and so on and it is a compression wave that is important and we will discuss that in greater detail as we go along and the compression process across the shock wave is highly irreversible and so and as we as I mentioned in the previous lecture any irreversibility flow is tantamount or any loss of stagnation pressure is tantamount to irreversibility and irreversibility as you know results in loss of exergy. So, this is undesirable in most cases although in some cases actually we can exploit the efficiency of normal shock compression process. So, you can actually think of normal shock as a means of compressing a flow without any moving parts. So, we do not utilize a compressor, you can actually compress a flow by means of a normal shock. So, when you look at it that way for certain sorts of examples we can actually exploit this fact, it is very desirable, we will take a look at these things as we go along. Now if you really think about physical situations where one would encounter this for example in addition to this, for example normal shock wave is rather like a blast wave you know which goes across it is a single wave very powerful and moves at the wave itself moves with supersonic speed if you are an observer standing like this and looking at the wave. So, ahead of the wave is a quiescent air for instance and behind the wave is actually air that has been affected by the passage of the shock and in the case of the shock wave the air behind it moves with very high speeds as well unlike that of an acoustic wave and the wave itself moves with the supersonic speed. Now this is in contrast to an acoustic wave which is actually a wave train. So, it is a it is alternating compression and rarifaction as the wave goes through. So, that is the big difference between an acoustic wave and a normal shock wave. Normal shock wave is not a it is not a wave train it is a single wave front whereas an acoustic wave is usually composed of wave fronts with compression and rarifaction. Now, we use the term normal shock wave and the term normal is used to denote the fact that the shock wave is normal or perpendicular to the flow direction. Just like what we showed for the for the acoustic wave in our view of things for example, in a frame of reference where the nozzle I am sorry where the shock wave is stationary the direction of normal shock wave is perpendicular to the flow direction as you can see here oops as you can see here and there is no change in the flow direction as a result of passing through a normal shock wave. So, in this case the flow is perpendicular to the normal shock wave as it approaches the shock wave and it remains perpendicular to the normal shock wave after it passes through the shock wave. Now, this is in contrast to oblique shock waves which are more frequently seen in nature where the flow is deflected or turned as after it passes through the shock wave it may be perpendicular approaching the shock wave, but after it passes through the shock wave it actually is inclined to the shock wave. So, we will now take a look at take a look at thermodynamic and flow aspects of normal shock waves in calorically perfect gases only. And the theory is far too complicated in the case of steam or refrigerant, so we will not do that we will only look at the thermodynamic and flow aspects of normal shock waves in calorically perfect gases. So, in a frame of reference where the shock wave is stationary and the situation is as depicted here more or less similar to what we had earlier for an acoustic wave. The only difference being let me just show that the slightly different color, so in the case of an acoustic wave, so this is 1, this is 2, so the speed of the fluid as it approaches the wave is equal to the speed of sound and the speed of the fluid after it passes through if you recall we said this was V2 equal to V1 plus dV1, so it changes only by an infinitesimal amount whereas in this case as we have already said the speed is usually supersonic, so this is actually greater than the speed of sound corresponding to this state A1, so if you want to be very specific we can say this is equal to A1. And V2 is also not infinitesimally different from V1, it is actually considerably different from V1. So, we have mentioned this and changes in properties across an acoustic wave are infinitesimal and hence isentropic whereas they are large and irreversible across a normal shock wave, so which means that S2 is greater than S1 in this case and here S2 is equal to S1. So, once again we write the governing equations in the usual form continuity momentum and energy equation and by using the definition of the Mach number and calorically perfect assumption we can come up with expressions like this. In fact, if we combine all the three expressions we end up with something that looks like this. Notice that M1 is known because that is the speed of the fluid approaching the shock wave, so given P1, T1 and V1, so those are known. So, in this shock wave if you see T1, T1, V1, so this completely describes the state of the fluid before the shock wave. So, given these quantities how do we determine V2, P2 and T2, so that is the question that lies before us. So, that means M1 is known, so M1 is known and M2 is not, M2 is unknown. So, although this equation looks somewhat complicated it is actually, it actually is not, it is in fact quadratic equation and M2 square. So, in fact if you go ahead and solve this quadratic equation you can do so in closed form and obtain an expression for M2 in terms of other quantities. Notice that once M2 is known all the other properties may be evaluated say using this relationship and using this relationship. M1 is known if M2 is also known then I can get P2 and T2. V2 can also be obtained once M2 is when these quantities are known. So, if you solve this you get only one meaningful solution to this other solutions are M1 equal to M2 and so on which may be neglected. So, we get this expression for M2. Now, this actually admits two solution one for which M1 is greater than one the flow is supersonic approaching the wave and M2 being subsonic. So, after passing through the shock wave the flow becomes subsonic and S2 is greater than S1. So, this is a compression solution that is very, very important this is a compression solution. So, P2 is greater than P1 and in fact T2 is also greater than T1 in this case. The other solution that is admitted by mathematics is that M1 is subsonic and M2 is supersonic. So, the flow is accelerated to supersonic speeds after passing through the shock wave and S2 is less than S1. Since this is an adiabatic flow remember our energy equation looks like this. So, there is this corresponds to an adiabatic flow Q is equal to 0. So, that means that the entropy must remain the same or increase in an adiabatic flow. So, S2 less than S1 is not possible this is forbidden. So, this solution is not permitted or not seen in real life although mathematics allows it in real life this is not seen because it would violate second law of thermodynamics. Because one more solution that is possible is simply that of an acoustic wave M1 equal to 1 M2 equal to 1 and S2 equal to S1. So, this is isentropic process with the Mach number approaching the wave being equal to 1. So, this is the solution that we will focus on. Now, let us try to show this solution normal shock solution on a TS diagram. So, on a TS and a PV diagram. So, state 1 P1 T1 and V1 are given. So, you know that this is equal to 2 Cp and this is the stagnation state corresponding to state 1 0 1. So, T0 and P0 for this stagnation state are also indicated here and the same information is shown on a PV diagram here. So, this is this is S equal to constant line state 1 lies here. This is the isotherm corresponding to the stagnation temperature. So, the point of intersection of the 2 is the stagnation state and this is P01. So, the point of intersection of the S equal to constant line and the stagnation temperature line gives us the stagnation state. Same thing is shown here also. Now, remember this is a solution for which M1 is greater than 1. So, we need to show that explicitly here which we have done here. So, once I have T0, I can calculate T star in the same manner as I showed in the previous lecture. So, the acoustic state star state is shown here. So, this is the M equal to 1 line all states that lie below this or supersonic states and all states that lie above this line or subsonic states. So, clearly state 1 is a supersonic state. We have done the same thing. So, we have drawn the isotherm corresponding to T equal to T star here and the line of and the point of intersection of the T equal to T star isotherm and the S equal to S1 line uses the sonic state. Clearly, this is a supersonic state. So, now we wish to locate the downstream state on this diagram. Remember the important things for the downstream state or that M2 should be less than 1. So, that means it is a subsonic and S2 should be greater than S1, which means that state 2 since S2 is greater than S1, state 2 lies to the right of state 1 and since it is subsonic, it should lie above the M equal to 1 line. Now, bear in mind that T0 itself remains constant because the flow is adiabatic. So, as you can see from here, so which means that T star will remain the same. So, this is always the M equal to 1 line because T0 does not change. So, that means state 2 should or probably will lie somewhere like this. So, that S2 is greater than S1 and M2 is less than 1 and in this case, this is S equal to S1. Let me show it here like this. So, this is S equal to S1 and we now draw a new line S equal to S2. So, this is S equal to S2 and again the state is subsonic which means that it will lie above this. So, state 2 will most likely be somewhere there. Let us see. So, state 2 is shown here. This is P2, T2 and remember T0 is the same. So, you can see that T0 is the same. So, the line of intersection point of intersection of S equal to S2 equal to constant line and the stagnation temperature gives us the stagnation state. Same is true here. This is S equal to S2 and this is S equal to S1. So, the point of intersection of S equal to S2 and the isotherm T equal to T0 gives us the stagnation state and this is P02. So, you can clearly see from the PV diagram that P02 is less than P01 which means that there is a loss of stagnation pressure as a result of the irreversibility S2 greater than S1. So, we have now we are now encountering a loss of stagnation pressure which is a concomitant of the normal shock compression process. What is that? We have connected state 1 and 2 with this line but there is really no line. So, it is a dashed line and you must be clear that it is a wave solution which means we have state 1, we have state 2. The process there is no process connecting state 1 and state 2. It is an irreversible process and we do not know how the two are connected. So, we are treating it as a wave like solution which means it is discontinuous. So, the flow jumps from. So, if you actually look at the illustration here, the flow jumps from state 1 to state 2. In fact, if you actually again think about the acoustic wave and the normal shock wave. In the case of an acoustic wave, we get a series of a train of compression and rarifaction fronts and so when I hold up let us say let us say a pitot tube like this where I want to measure the static pressure and the sound wave actually passes over the pitot tube. The process undergone by the fluid as we measure its pressure is isentropic which means the pressure disturbance increases or decreases the pressure only slightly. So, if I try to sketch that on a PV diagram like this, so let us say this is our S equal to S1 line. So, what would happen is that you would see something like this. So, the pressure would increase or decrease like this by a small amount. I have exaggerated that greatly here. So, you are going to see a fluctuation of the pressure about a mean value both on the positive side and negative side as the train of compression and rarifaction front passes through the pressure just goes up and down like this as I have shown here. Now, in the case of a normal shock, it is a single wave which passes through. So, the fluid would be at some state and then as the flow as the wave passes over this, it would jump to a higher pressure just in one go and then the pressure will remain the same for some time and then it will slowly relax back to the ambient condition. So, it would not look like this. It will probably jump to a higher pressure like this from here to here and then it will slowly relax over some time and then come back to the atmospheric value or what are ambient pressure value. So, that is the difference between an acoustic wave solution and a normal shock solution. So, this is in a slightly different frame of reference. So, in case you find it confusing, we can always do this in a stationary frame of reference where the wave is moving. Remember, this is in a moving frame of reference. So, let us do it like this. So, let us say that this is P ambient, this is the initial state beneath my probe. So, this is P ambient and it is at let us say T ambient. So, this is the initial S equal to S1 line and let us say that this is T equal to T ambient. So, now when an acoustic wave passes through my probe, we are in a stationary frame of reference. So, the acoustic wave comes through here alternating compression on dry refaction waves. So, what happens is and the process is isentropic. So, I move along this isentrope, I am sorry. So, I move along this isentrope like this, up and down like this. So, the pressure increase or decrease is infinitesimally small. Now, in the case of a normal shock wave, what would happen is we start from here and we jump to a the pitot probe shows an instantaneous increase in pressure and temperature like this. And then it slowly relaxes back to the ambient temperature. So, it slowly relaxes back to the ambient temperature. And then it slowly comes back to the original state that it was in. So, it will jumps like this and then it comes back like this. I suggest that you consult the textbook. I have given this in great detail in the textbook. So, please look at it if you are interested. So, that is the difference between an acoustic wave and a normal shock wave. So, what I have done here is what I have done next is, so we have this solution M2 and I have plotted all these quantities P2 over P1, T2 over P1, P02 over P01 as a function of the initial Mach number. So, let us take a look. So, this is done both for diatomic and monatomic gases. Although we will probably not discuss that aspect too much in this course, we will only take a look at the qualitative trends. So, as you can see, as the initial Mach number increases, the Mach number after the flow, after passing through the after passage through the shock wave keeps getting smaller and smaller and it sort of reach, it seems to reach an asymptotic value. It is not going down to like 0 or anything like that. It seems to be reaching an asymptotic value which we will show in a minute.