 The reason I want to talk about shock waves here is because we are in the process of discussing these wave kind of phenomena and the only difference between a sound wave and a shock wave is that we relax that condition that across that wave front, if it is sound wave we are dealing with infinitesimally small amounts of change for pressure density etc. If we relax that assumption or condition what you end up generating is essentially what we call a shock wave. So, a shock wave is nothing but again a wave front across which fluid properties like temperature, pressure density etc. will change but by finite amount. So, that is the only difference between a sound wave front and a shock wave front. In fact many people will point out that a sound wave is essentially a special case of a shock wave where the changes across the wave front are infinitesimally small. The important point though that one may want to discuss a little bit although technically we do not discuss this in a thermodynamics course much is the formation of a shock wave. So, I have just put together one single slide to talk about the mechanism with which a shock form. We are going to talk about a shock formation in our simplified one dimensional situation and the way you want to perhaps explain to your students what is the mechanism of a shock formation is that if it turns out that in some situation there is a formation of a sequence of sound wave one after the other each of which increases the pressure, temperature and density by a small amount. Then what ends up happening is that after a certain small amount of time we end up seeing a what is called as a coalescence or merging of these individual sound waves into one single wave front across which the properties change by finite amount and that single wave front is essentially what is called as a shock wave front. So, roughly speaking you can imagine that we are talking about again that explosive or a strong firecracker let us say going at a location. What again is happening once that explosive goes off is a certain amount of energy is getting dumped or getting deposited at the location where it is going off. Corresponding to this deposition of energy the pressure, the temperature, the density etc are all increasing at that location let us say. So, you can always conceptualize this deposition of the energy process and the consequent increase in pressure, temperature density as being carried by a sequence of individual sound waves and that is precisely what I have shown on my slide here. So, what we conceptualize the situation here is that a large number of individual sound waves is formed. So, the first sound wave let us say is moving with a speed of A as it was discussed earlier. So, ahead of this first sound wave we have gas medium which is stationary let us say. What we said is that there is a sequence of these sound waves. So, across the first sound wave there is already an increase in temperature through the amount delta t let us say. So, what we have is t in front of it and t plus delta t. So, let me go back to my board and let me draw this little more detail wise. So, I have A here p, t, rho etc, v is equal to 0. Then I have the second wave front which now is not moving in a temperature of t, but it is moving in a temperature of t plus delta t likewise there is p plus delta p and so on. So, consequently the speed with which the second sound wave front is going to move is going to correspond to an elevated temperature t plus delta t and therefore, this speed is going to be slightly higher than the speed of the first sound wave front which was moving with the speed of A. And this process then continues backwards and if you want to imagine successively later sound wave fronts each of these guys is moving faster and faster with respect to the previous wave front and that is precisely what I have drawn on my slide. So, going back to my slide again what I have shown here is a sequence of sound wave fronts. The first one is moving with speed A, the next one which is moving behind it is at an elevated speed A plus delta A or D A and likewise the successive sound wave fronts are going to be moving faster and faster each of which has a certain delta p delta t delta rho etcetera associated with it. Since each successive sound wave moves faster than its predecessor what ends up happening is that eventually the latter wave will actually overtake the former ones and invariably what is found to happen is that all of these waves will merge together so that a thin wave front is formed across which the properties now will not change by infinitesimally small amount, but will change by finite amount and this is what we call a shock wave front. So, formally this is the mechanism with which a shock wave form. We are talking about this situation in a truly unsteady situation in the sense that we are sitting as a laboratory coordinate system and we are seeing these sequence of sound waves getting generated one after the other and overtaking one another to finally form a single shock wave front. Shock waves do occur in steady flows also and the mechanism with which they occur in a steady flow situation is no different than what I am talking about here. So, finally when it comes to formation of a shock wave if you have to explain it to your students this is the mechanism with which you can explain it that it is the coalescence or merging of successively faster moving sound wave front that results in a shock wave front. At this time you know when it comes to a thermodynamics course where you are discussing these compressible flow ideas. I do not expect and anticipate any more explanation than these normally a far more rigorous analysis can be performed and we do do that in case of a compressible flow type course, but I think purely based on a physical description if you want to convey what is really causing a shock formation this is the mechanism with which you can explain it. So, then finally to summarize what we have been talking about the shock wave is essentially a wave front across which fluid properties change by finite amount and physical measurements actual experimental measurements have shown that the stream wise extent or the extent along the flow of this so called shock wave region is very very small and typically it is of the order of 10 to the power minus 7 meter. So, it is a tenth of a micrometer in standard conditions what I mean by standard conditions is roughly the atmospheric pressure and atmospheric temperature into which a shock is formed let us say. So, as far as an analysis is concerned we do not bother about what is happening through that small distance of 10 to the power minus 7 meter. We simply say that let us treat this shock as a discontinuity we form a shock wave front just like what we did for the sound wave front we attach a coordinate system to the shock wave front we draw a control volume around the shock wave front just on either side. So, that the inlet to the control volume or the open system is on one side of the shock wave front the outlet is on the other side and what we just wrote here is that the properties of the fluid change by finite amount across the shock. Therefore, let us say that with respect to this coordinate system which is attached to the shock wave front the flow ahead of the shock which is coming toward the control volume or the shock wave front be at a speed of v 1 corresponding to which there is a mach number m 1 and the pressure temperature and density of the flow ahead of the shock wave front be p 1 rho 1 and t 1 etcetera. Let us say that all these upstream conditions as they are called with respect to the shock front are known and the formal objective of the shock wave analysis is that knowing p 1 rho 1 t 1 v 1 m 1 etcetera I want to find out p 2 rho 2 t 2 v 2 m 2 which is basically the set of conditions behind the shock wave front. The analysis is no different than what we did in case of the sound wave front the only good part of the sound wave front was that because the changes were infinitesimal we could neglect certain terms and we could simplify the expressions to obtain the speed of sound front. Here on the other hand we cannot neglect anything because fluid properties are changing by finite amount across the wave front. So, in this case what is done is with respect to this control volume we end up employing our standard governing equations of motion which are nothing but balance statements for mass linear momentum and energy and that is precisely what I have written. So, let me go back to my board and let me draw this once. So, this is my shock wave front I have attached my coordinate system and the control volume just around the shock wave front and I have conditions 1 which are known and conditions 2 which are to be determined and what we end up doing now is the coordinate system is attached to this shock wave front just how it was in case of sound wave front. So, therefore with respect to this coordinate system the flow is steady. What this coordinate system sees is that the flow is approaching the control volume or the open system with a velocity v 1 and is leaving the control volume or the open system with a velocity v 2. All other conditions are rho 1, p 1, t 1 and rho 2, p 2, t 2 you can add that Mach number because Mach number is simply now going to be the ratio of this velocity v 1 to the speed of sound a 1 which can be found once I know the temperature t 1. So, m 1 will be simply equal to v 1 divided by a 1 which is equal to v 1 divided by square root gamma times r times t 1. So, you can add that m 2 here if you want and that is going to be equal to v 2 divided by a 2 which is v 2 divided by square root of gamma r t 2. So, that is what I have outlined on my slide earlier that known conditions in 1 we need to find out conditions in 2 and to do that all we need to do is employ our steady flow governing equations. So, the first one which is equation number 1 is simply m dot in equal to m dot out as far as that control volume is concerned that is the mass balance equation which is obvious. The second one here equation number 2 p 1 plus rho 1 v 1 squared is equal to p 2 plus rho 2 v 2 squared is essentially your momentum balance. I want you to go back and make sure that with respect to the control volume that we drew here and the momentum balance statement which I had written right at the beginning of this discussion make sure that this is indeed the momentum balance that comes out. So, p 1 plus rho 1 v 1 squared is equal to p 2 plus rho 2 v 2 squared and the third equation is your steady flow energy equation again note here that for the control volume there is no external heat transfer and there is no external work transfer noting this we can immediately simplify our one dimensional steady flow energy equation in the form of h 1 which is enthalpy in conditions 1 plus v 1 squared over 2 equal to h 2 plus v 2 squared over 2 and we are dealing with ideal gases here. So, I am going to replace the enthalpy in terms of this product of the specific heat times the temperature. I am not talking about here a reference temperature as such and that is perfectly fine I have essentially defined a reference temperature at some suitable low value with respect to that I am calculating these enthalpy values and remember finally we are going to always require changes in the enthalpy. So, that reference temperature or the reference enthalpy if you want is going to cancel out from this subtraction. So, without loss of any generality then I am in a position to write h 1 as C p times T 1 and h 2 as C p times T 2. So, here obviously I am assuming that the constant the specific heat at constant pressure is essentially a constant and that is what the assumption within which we are working here. Additionally, we note that both in conditions 1 and 2 ahead of the shock and behind the shock the ideal gas equation of state will apply. So, therefore p 1 is going to be equal to rho 1 R T 1 and p 2 is going to be equal to rho 2 R T 2 and this is really it as far as performing the normal shock analysis is concerned this set of equations is what you are going to manipulate to eventually obtain what we want. So, the remaining part is simply algebra there is nothing more than that once you understand where these equations are coming from namely steady flow mass balance, steady flow momentum balance and steady flow one dimensional energy equation along with the equation of state applicable on both sides of the shock wave front we are in a position to perform the algebra and find out the final solution. So, what I have done here is I have outlined the steps using which you can perform this algebra. So, for example, what you can do now here and by the way this is one way of doing this analysis there are couple of different ways using which you can perform this normal shock analysis I am outline outlining one such procedure depending on the book that you open you will see that authors follow different procedure. Obviously, the final answers are the same, but I am outlining one such procedure here for the normal shock analysis. So, going back to my slide what you can do is using equation number one which is our continuity equation as it is called in fluid mechanics or mass balance I can eliminate v 2 squared in this momentum balance and with a little bit of algebra you can rearrange the momentum equation in the form that is written out here. So, my suggestion and request is that go back at the end of this lecture and make sure that you are in a position to obtain these intermediate results using what has been suggested that using equation number one eliminate v 2 squared in equation number two to simplify equation number two to obtain and slightly modified momentum equation. So, that is what I am going to call equation number five which is an intermediate result. One more thing that you can do here is again using this equation number one which is our mass balance you can eliminate v 2 squared over 2 or v 2 squared I should say in the energy balance that is in equation number three and which is what I have written out here and you can reduce that energy equation mind you this requires reasonable amount of algebra. So, the previous equation simplifying the momentum equation to something like this using the continuity did not require too much algebra as you can verify when you go back. On the other hand eliminating that v 2 squared in the energy equation using the mass balance and obtaining this equation number six which I have written out here does involve reasonable amount of algebra. However, it is worthwhile doing it once so that you know exactly how these things are obtained. Finally, going back to equation number five which was a slightly modified momentum equation we make sure that here there is that v 1 squared which also appears here in equation number six which is a slightly modified energy equation. So, then finally using that modified momentum equation you can eliminate this v 1 squared in the energy equation and simplify again reasonable amount of algebra to finally obtain the relation which is now being shown as equation number seven which is basically the ratio of densities across the shock wave front rho 2 divided by rho 1 which is also by the way equal to v 1 over v 2. Why? Just go back to equation number one rho 1 v 1 is equal to rho 2 v 2 through our mass balance. So, therefore rho 1 v 1 is equal to rho 2 v 2. So, rho 2 over rho 1 then is equal to v 1 over v 2 can be expressed as a unique function of the ratio p 2 over p 1 and the specific heats ratio gamma. So, equation number seven is one important result in this normal shock analysis. Once you obtain equation number seven what you can do is you can utilize these equations of state relations which are our equations number four and using equations number four here along with equation number seven do a little bit more algebra and you will express the ratio of the temperatures t 2 to t 1 uniquely again in terms of the ratio of pressures p 2 to p 1 across the shock again that gamma will show up in this expression as well that is perfectly fine. So, finally the important results of this normal shock analysis are equation number seven which was shown on the previous slide and equation number eight which is shown on the present slide and these two together is what is popularly called as the Rankine-Ugonio relations. The reason is because Rankine and Ugonio were the people who perform this analysis and came out with these expressions. What we are supposed to interpret from these relations is that the ratio of the densities hence the ratio of the velocities and the ratio of the temperatures across the shock wave front can be uniquely expressed in terms of the ratio of the pressures across the shock wave front which is given by this p 2 over p 1 and the ratio of the specific heat at constant pressure to that at constant volume which is gamma. And therefore, to summarize this normal shock analysis the Rankine-Ugonio relations that we just obtained are basically what we call a normal shock solution in terms of the parameter p 2 over p 1 because remember going back to those two expressions you see that once you know the ratio p 2 over p 1 and gamma you are in a position to evaluate t 2 over t 1 rho 2 over rho 1 and v 2 over v 1 etcetera. So, the Rankine-Ugonio relations then are the normal shock solutions in terms of the parameter p 2 over p 1 and this parameter is what is called as the shock strength. So, this is one of the most important parameters in a normal shock analysis p 2 over p 1 and that is what is called as a normal shock strength. So, far what we have been up to is basically algebra. We have manipulated a certain set of equations, but those equations have resulted into a set of final relations which we are calling Rankine-Ugonio relations through essentially algebra. When we want to include concepts of thermodynamics in this normal shock analysis what we will stipulate is that because the situation is such going back to my control volume that there is no external heat transfer in this control volume for the process to be possible entropy in station 2 has to be greater or equal to entropy in station 1 that is what will be mandated or required by the second law. So, we write an expression for the change in the entropy let us say specific entropy across the shock wave front as S 2 minus S 1 using the standard expression C p times the logarithm of t 2 over t 1 minus r times the logarithm of p 2 over p 1 and what can be done here which I am not doing as algebraic exercise is this t 2 over t 1 ratio we can express using our previously determined Rankine-Ugonio relation in terms of p 2 over p 1. So, that this entire S 2 minus S 1 which has to be greater than or equal to 0 as required by the second law expression can be couched in terms of a big expression involving the shock pressure ratio or the shock strength p 2 over p 1 and for S 2 minus S 1 to be greater or equal to 0 you can show that this p 2 over p 1 has to be greater than or equal to 1. My request is that you have all the relations that are required to perform this algebraic exercise. This t 2 over t 1 expression as a function of p 2 over p 1 and gamma is available here. So, my suggestion and request is that go to this expression substituted for this t 2 over t 1 simplified and finally, enforce the condition that from the second law of thermodynamics S 2 minus S 1 has to be greater than or equal to 0 for this shock process to be feasible and in doing so you will come up with the result that the shock pressure ratio or the shock strength as it is called has to be greater than or equal to 1. In other words what we come to the conclusion here is that the pressure and therefore, you can quickly conclude that the temperature and the density as well must increase across the shock and correspondingly the flow velocity must decrease across the shock. So, this is what we will formally call as the role of thermodynamics in the shock process that it demands or it mandates that the shock pressure ratio or the shock strength has to be greater than or equal to 1. This is more or less really what the normal shock analysis is. Let me talk about quickly an alternative manner in which you can perform this normal shock analysis. So, what we have seen so far in the last half an hour or so is the Rankine-Ugonio relations which is or which are essentially the normal shock solution in the form of that shock strength P 2 over P 1. Instead of using that shock strength P 2 over P 1 as the parameter some people prefer to express these solutions in terms of the Mach number M 1 as the parameter. So, for example, although I am not doing the complete algebra here you can uniquely express the shock pressure ratio or the shock strength P 2 over P 1 equal to or in terms of the M 1 which is the Mach number of the upstream flow which is simply V 1 divided by square root of gamma R T 1. And once you uniquely express P 2 over P 1 in terms of M 1 and gamma all these previous expressions such as rho 2 over rho 1 V 2 over V 1 or V 1 over V 2 and P 2 over P 1 can be expressed rather than P 2 over P 1 in terms of M 1 and gamma. So, let me just quickly show why this can be done although I am not doing this completely. Let me go to the board that I have. So, if you write the momentum equation or the momentum balance I would say this is what we had and let me rewrite this as minus rho 2 V 2 squared on the right hand side I will take rho 1 V 1 squared as a common factor. So, 1 minus now rho 2 V 2 squared I will write it as rho 2 V 2 times V 2 divided by rho 1 V 2 divided by V 1 times V 1 and rho 2 V 2 is exactly equal to rho 1 V 1 through our mass balance. Therefore, what I have is rho 1 V 1 squared times 1 minus V 2 over V 1. Now if I on the left hand side divide by P 1 I can write this as P 2 over P 1 minus 1 and therefore I will have to divide this right hand side also by P 1 and it is very easy to show that this ratio rho 1 times P 1 squared divided by P 1 is exactly equal to gamma times M 1 squared. So, are going to the next page let me just rewrite this P 2 over P 1 and equal to 1 plus gamma times M 1 squared I will have 1 minus this V 2 over V 1 right. So, now what we can see is that this V 2 over V 1 ratio can be expressed uniquely in terms of P 2 over P 1 the shock strength and gamma we already have P 2 over P 1 on the left hand side here. What this means is that this ratio P 2 over P 1 which we have been calling a shock strength can be uniquely expressed as some function of M 1 and gamma. So, instead of using this shock strength P 2 over P 1 as the primary parameter some people will prefer to use the upstream Mach number M 1 as the primary parameter and express the shock solution in terms of M 1 rather than P 2 over P 1. So, going back to my slide then you can actually show that then this P 2 over P 1 is equal to 1 plus 2 gamma time over gamma plus 1 times M squared minus 1 etcetera. Here again if you want to invoke your second law of thermodynamics which will demand that S 2 minus S 1 has to be greater than or equal to 0. We can show that if this has to be true then M 1 which is the Mach number of the flow with respect to the coordinate system that is attached to the shock that is ahead of the shock has to be greater than or equal to 1 and immediately then as a consequence we can show that M 2 which is the Mach number of the flow which is behind the shock front has to be less than or equal to 1. So, to summarize the upstream flow with respect to the shock must be supersonic and the downstream flow with respect to the shock must be subsonic. And just to conclude this discussion here what we saw here is that the shock strength is uniquely expressed in terms of upstream Mach number M 1 and gamma. Likewise all property ratios T 2 over T 1 rho 2 over rho 1 V 2 over V 1 all of those can be expressed again as some unique functions of M 1 and gamma and those functions are available as a tabulated form and that is what we call a normal shock table. So, for a given gamma we can choose the gamma equal to say 1.4 you will see that the normal shock table will typically include these columns. So, M 1 which will necessarily start from 1 because only for an M 1 of greater than or equal to 1 we have a physically means meaningful process of the normal shock. So, corresponding to M 1 of 1 M 2 is also exactly equal to 1 and then all these ratios T 2 over P 1 rho 2 over rho 1 T 2 over T 1 and a few other ratios which we will talk tomorrow little more are tabulated as functions of M 1. So, with this what I will do is I will stop for today.