 Good morning and welcome you all to this session of the course. Now, last class we were discussing about the oblique shock wave and where we ended just we have to look that last class we are discussing again we just make a recapitulation that a oblique shock wave is like this where the inlet velocity or the incoming velocity v 1 is being deflected and this angle is the angle of deflection and this is beta and what we discussed last class is that this oblique shock wave can be treated as a normal shock provided the Mach number Mach 1 which is v 1 by a 1 is modified at m 1 sin beta and the Mach 2 actual Mach number is v 2 by a 2 the condition here took the section 2 section 1. So, it is m 2 that is the normal component sin beta minus delta that is this angle made by this velocity v 2 with the line parallel to the shock wave and similarly this is beta. So, this we discussed and finally, we under appreciated that a relationship between beta and delta and the Mach number initial m 1 sin beta or m 1 whatever you tell is very much important and then from a trigonometric relationship we arrived at this relation tan delta is 1 minus x where x was some quantity defined. So, finally, this is the expression this is the expression we derived where delta is this deflection angle beta is this shock wave angle well m 1 is the approach Mach number m 1. So, this is a very important relationship now two characteristic feature of this relationship is that you see when beta is pi by 2 this already we discussed earlier beta pi by 2 means that when the velocity of approach to the shock becomes 90 degree makes 90 degree angle to the shock that means this becomes a normal shock this is the normal shock condition normal shock then what happens what is the value of delta when beta cot beta will be 0 when beta is pi by 2 cos cot beta will be 0. So, therefore, delta is 0 now another case you see when m 1 sin beta is 1 then also delta is 0 this is because this becomes 0 and this m 1 sin beta 1 means sin beta is 1 by m 1 and this is the Mach angle that means beta becomes equal to the Mach angle because we know that for a Mach wave the angle which it makes with the flow stream direction is 1 by m 1 sin of that that means sin inverse 1 by m 1. That means this beta becomes the Mach angle that means the shock wave becomes a Mach wave. So, therefore, two extreme conditions where normal shock we already appreciated that the value of beta lies between two extreme values pi by 2 and sin inverse 1 by m 1 that means here you can write beta is equal to sin inverse 1 by m 1. So, within these two limits there is no deviation this is because for a normal shock the velocity before the shock and after the shock at the same direction similarly for a Mach wave the velocity before and after the shock the same and this again can be represented by this like this that a oblique shock is that oblique shock is between this two limit one is normal shock normal shock and another is the Mach this is the Mach wave Mach wave here also there is no deflection V 1 V 2 and the oblique shock remains somewhere here where there is a deflection of the velocity that is the delta sorry you cannot understand delta. So, therefore, the oblique shock wave lies between these two limit now you see that this particular formula this can be little bit we can do with this since beta is delta is 0 between pi by 2 and sin inverse 1 by m 1 which means that beta may have either maximum or minimum in between these two value because it is 0 0 to 0. So, that can be found out by making a derivative of this that if we can make this with respect to beta d delta d beta making 0 and checking for the second derivative whether it is less than 0 or greater than 0 we can find out the condition of maximum and minimum and if you perform this then this is your task I am not doing this the entire algebra in the class, but I tell you this thing if you do it then you get the delta has a maximum that d delta d beta is 0 when beta is given by this expression sin square beta. Now, this beta well in the book they sometimes write the max, but I will write the optimum value of beta where it is maximum delta that is gamma plus 1 by 4 gamma this is relatively complicated expression 1 by gamma m 1 square into 1 minus you can do it at your home this is a task for you 1 plus gamma minus 1 by 2 m 1 square plus gamma plus 1 divided by 16 m 1 4 really a little complicated one. So, this is the expression for the optimum value of beta where delta is maximum and if you put this value and you get the maximum value of delta that expression you can find out and more important thing is that how does it look now if you make that you will can find out the value of delta maximum and that all depends upon the mach number. So, therefore, the optimum value of beta where delta is maximum and the maximum value of delta which will be found out by replacing this in this equation for the value of beta one can get it is not very simple, but straight forward thing it will complicated. So, one can get a curve like this that I will explain now what is this curve we see it very carefully the curve is like this now we can find out we see that from here that now we what we are doing we are drawing the deflection angle with beta or beta with delta for different values of m 1. So, a family of curves for delta beta with different values of m 1 now for a given value of m 1 let us understand the curve as we have seen that beta is 0 here where sorry delta is 0 here where beta is sin inverse 1 by m or m 1. So, this is for a particular value of the m 1 this is for a given value of m 1 clear and it has a maximum that is the maximum delta that is the delta maximum clear now this as a value like that and this what is this this is the pi by 2 that means this is the normal shock limit this is the max. So, this is the shock wave angle for a given value of m 1 or all right and this clearly has a maximum value that is delta max here another one curve is drawn this one this is this is this is drawn because this gives the value of the beta optimum now what is this is the locus of m 2 now if you find out the locus of m 2 in the beta and delta plane just like that this is the locus of a given m 1 that how beta and delta changes for a given m 1 we know that the m 1 is related with m 2 this m 1 is related with m 2 this m 1 is related with m 2 because m 1 sin beta earlier we did that m 2 sin beta minus delta they work as the max number approach or inlet or at the upstream for a normal shock and that is after the normal shock the way the normal shock relates the downstream max number to the upstream max number with the same relation we have shown that is valid for this case. So, therefore, if we do that mathematics then we can plot in the figure the locus of the constant m 2 that means this is one and this is done for m 2 is equal to 1 I will explain it after what this is done for m 2 is equal to 1 m 2 is 1 means m 2 is the max number after the oblique shock. So, that locus is given. So, this is the picture now another interesting picture which is found from this figure is that for a given value of delta that is the deflection angle we get two values of beta that means there are two values of beta solution is obtained. So, now why abscissa is not coming given value sorry given value of delta there are two values now it is it is it is it is focused already it is ok. So, for a given values of delta we have two values of the shock angle now this delta is specified by the geometry of the flow this delta is the turning angle that we will understand solving problems that delta depends upon the geometry of the problem because delta is the turning. So, shock wave will take place depending upon the turning of the flow. So, the flow is turned by the geometry of the flow for example, the flow passes through a wedge. So, therefore, this deflection angle in fact is the input to the real problem. So, therefore, for a given delta depending upon the geometry of the problem we get two solutions from this as we see that two beta value. Now, if this two solutions if we see we will find out that you see this that is why a mach number one curve the locus of mach number one is given here m 2 1 mach number means m 2 1 and the values of m 2 below this curve in this plane is m 2 less than 1 and the value of the sorry m 2 less than 1 I am sorry it is m 2 greater than 1 it is already given m 2 greater than 1 and above this we can get the locus of all constant m which is m 2 less than 1. That means this solution of this shock wave angle gives a shock where after the oblique shock the velocity that is m 2 is greater than 1 the flow remains in the supersonic that already we discussed that after the oblique shock the flow may be supersonic may be sonic because the restriction is on the m 2 sin beta minus delta which has to be less than 1. So, m 2 may be greater than 1 whereas, this solution that this value of shock angle that the higher value gives rise to m 2 less than 1. That means if we read the m 2 value here that means this is a strong shock because it reduces the velocity to subsonic level and this is the weak shock. So, therefore, two solutions of beta out of which one the lower value corresponds to a weak shock solution and the higher value corresponds to a strong shock strong shock solution. So, usually it is very difficult it depends upon the circumstances which one will prevail, but it has been found experimentally that flow past external bodies or the external flows the solution or the shock wave the weak shock wave is found that means after the shock the flow becomes supersonic, but it may be subsonic depending upon the circumstances. So, if nothing is told for our purpose for solving problems specifically we usually take the solution of weak shock that we will do when we will solving the problem, but this is the total physical picture that the shock wave angle dependence with the turning angle for different values of m 1 and at the same time what will be the corresponding values of m 2 that is the mach number after the shock. Now, this thing can be shown here in a more compact form like this just you see this one is your this curve is your m 2 greater than 1. Now, you see this is a curve you can see probably this is deflection angle deflection angle delta this is deflection angle delta this is shock angle this is beta. Now, you see these are the families of curves for different m 1. So, this is one now one thing is that this is 1 by m 1 for given a for a given m 1 this is the for a given m 1 these are all corresponding mach wave limit, but the maximum limit is pi by 2 for all initial mach number. So, therefore, all this curve merges at 90 degree. So, m 1 values goes on this is m 1 values goes on increasing in this direction goes on increasing in this direction this is obvious, because if you see here the sin inverse 1 by m 1 is small when m 1 is high. So, therefore, m 1 increasing in this direction you see this is 1.4 1.6 234. Now, m 2 is equal to 1 this demarcates 2 series of 2 categories of mach numbers one category of mach number is here you can I will draw these are the mach numbers these are the mach numbers greater than 1 m 2 the less than 1 0.8 0.6 like that. Whereas, we get mach number here which are m 2 is greater than 1 that means, let for example 2. So, therefore, this curve this curve is m 2 1 m 2 1 demarcates the region of strong shock and the region of weak shock clear. And this one this this curve is the locus of the maximum deflection angle this is the locus of the this again coincides at the maximum value of the delta that I will come afterwards that delta max. Now, if what is this value why I stop here why the curve is like this is because if delta max is found out with m 1 then what is the value delta max by m 1 if you plot you will see that there is a limit. That means, if we now find just a minute I will show you that if we plot the delta max we calculate delta max and we plot delta max versus m then a interesting result will be there from 1 delta max. For example, delta max 0 10 this case I am just doing otherwise it will be easier 30 40 degree 50 then somewhere here for air it is calculated the values this value represents for air a for air a limiting value that 45.58 that means, with increasing m m 1 2 3 4 like that somewhere it is near 10 to the power 10 or something some value in ultimately asymptotic. That means, the maximum angle 45 that means, the turning angle cannot be more than 45.58 whatever may be the value of m whatever may be the value of m with m the turning angle changes that we have already seen the turning angle changes with m for a given shock wave angle you see we can have the turning angle changes with m with m changing the turning angle changes. But if we this is the curve if we now plot the delta max the maximum turning angle that means, this one this is the maximum turning angle for example, 1 m you choose this m m is equal to 2 this m curve what is the maximum turning angle this one this is the maximum turning angle if you choose this m this is the maximum turn that means, m is reduced if m increase this turning angle increases m is increasing in this direction I told you. So, as we increase the m the turning angle increase, but this becomes very close to each other and finally, this is totally become compact and finally, it reaches a value of 45.58 this is 45 45.58 for air this is a representative figure for air that means, if we increase the mach number very high also this will not reach go more than that then what is this that means, this physically implies one thing that if the angle is more than that just I am telling you two things coming to picture for example, the mach number is this one then I find out this is the maximum delta max mach number is this is the maximum turning angle for example, if mach number is 2 for example, I know this is the maximum turning angle if my turning angle is more than that what will happen do you know that if turning angle is more than the maximum that means, a flow through an wedge what is this turning angle turning angle is the flow is turned like this. So, if the wedge angle for example, this 2 alpha that means, the alpha is more than delta max for a given m 1 this wedge gives you the turning angle actually for flow then what happens for a given mach number the shock wave is not attached to this there becomes there occurs a detached shock this is known as detached shock detached shock the flow like this the detached shock is like this the flow takes place and this is a typical flow filled the detached shock this is m 2 is less than 1 this is m 1 and this becomes curved this is not this is not attached to the wedge this is detached shock this becomes curved and this type of shock is not amenable to the mathematical analysis we have done for a straight oblique shock. So, this is out of the scope of our discussion but you must know that this is there, but another interesting fact that if a body of this shape for example, of any shape which gives a turning angle given turning angle is accelerating with in the air that means, air flow is velocity is changing if it accelerates that it may so happen when the lower mach number is turning angle may be more than that, but higher mach number if we increase the mach number the turning angle increases when it accelerates and achieves a higher mach number then it may be such that the turning angle may be lower than delta max. So, therefore, when a body accelerates sometime it is observed that the shock is detached at lower mach number at again attached at higher mach number we will do we will solve some problems. So, that we can get some feel of this let us solve some problem then only you will let me see that I have some figure let me have a check that I have some figure which I will show like to show you this is the oblique shock is that shock curved shock detached shock. Let me solve a problem and then we will discuss the next part let us solve a problem first this is a first problem example one air flowing at m 2 mach number 2 with a pressure of the 80 kilo Pascal's and the temperature of 30 degree Celsius passes over a component of an aircraft that can be modeled as a wedge with an included angle of 8 degree that is aligned with the flow that is the flow is turn by both upper and lower surfaces of the wedge through an angle of 4 degree leading to the generation of a oblique shock wave find the pressure acting on the wedge circle just now I told that this is an example this type now what is this there is a wedge shape that means this is the wedge give a simple example the total angle included angle is 8 degree. That means this angle is 4 degree this side now this is the incoming air v 1 let the pressure p 1 temperature t 1 mach number is m 1 now what happens the flow will be turn through 4 degree both this sides that means one side we will just that means this is the flow which will be deflected because the streamline what will happen the streamline will be going like this so because of this wedge placed in the streamline is deflected and if the approach mach number is greater than one this deflection of streamline because of the presence of the wedge creates the oblique shock that is the concept that is the basic concept so this so this is the 4 degree this is the basic concept and this is the basic idea of the problem v 2 that is all so what we have to find out this is given that that is aligned with the flow that is the flow is turn by both upper and lower surfaces this is upper and lower surfaces of the wedge through an angle of 4 degree leading to the generation of oblique shock wave find the pressure acting on the wedge surface that means this surface p 2 so the same in both the sides I am just showing it here mach number will be m 2 temperature will be t 2 so how to solve the problem problem is very simple now m is 2 so first of all what we have to do with this m 2 the first job with this turning angle that means m 1 here is 2 turning angle here is 4 degree so with this m 1 with this turning angle we have to calculate the beta either from the equation you can tell sir I will calculate those who are very much interested in using the analytical equations that analytical expressions they can solve analytical expressions or you can see the chart because it will be easier that with m 1 2 and delta 4 if we see the chart m 1 2 we have to go for the different values this m 1 2 and delta at 4 degree one can find out what is the value of beta here beta also we will get 2 values one is for weak another is for strong and we take the strong solutions and from this figure we get the value of m 1 so from this what is the value of m 1 delta 4 m 1 sorry beta the value of beta is 33.4 degree that means we take the weak shock solution when nothing is told this is just a methodology weak if we can take strong shock solution also and find the result and which one will be correct that will be validated by the experiment so it is difficult to tell for the time being I am taking the weak shock solution so therefore the weak shock solution is the lower value of beta from the graph we are taking the beta 30 when beta is known things are done that means what is the procedure then we know m 1 sin beta we find out m 1 is 2 into sin 33.4 degree sorry and what is that value that is m n 1 that is m n that is that is told at m n 1 is m 1 sin beta and that become equal to 1.1 now we see the normal shock wave so from the normal shock wave with this as the m 1 plus plus I showed you that what the normal shock wave there will be an m 1 their column for in given m 1 we can find out the value of p 2 by p 1. So, for this m 1 this as the approach m to the shock or the inlet m 1 to the shock we can find out the value of p 2 by p 1 and that value is 1.245 that value is 1.245. So, this value if you just multiplied what is that 80 so p 2 is 1.245 into 80 kilo Pascal's so the value will be I am telling you the value I have done it it is 99.6 kilo Pascal's 99. Now, here this is the problem that pressure acting on the o h if it is told find out the velocity or the mach number here what is the mach number here I will find out the mach number after the flow how for this m 1 m n 1 from the shock table you find out m 2 that is not asked for, but I am adding this thing the book from where I have taken the problem this is not asked for now if I have to find out the m 2 here what you have to do you have to find out m 2 from shock table. That m 2 is what from shock table which you get from normal shock table is actually for the present case is m n 2 is actually m n 2 that is equal to m 2 that is m 2 from normal shock table is m n 2 and that is equal to m 2 sin beta. That means we know that for a oblique shock it is the sin component of the m act as the inlet and outlet m for a normal shock. So, when we solve the problem as equivalent to a normal shock always we have to read m 1 m 2 with respect to the normal shock wave which is nothing, but m n 1 that is m 1 sin beta and for m 2 it is sin beta minus. So, when we know both beta and delta we can from this we can find out the actual m 2 for this problem this is not given is not told here to find out. So, it is so this is one problem that we can discuss that for the shock then we will solve another problem then before that I think we have to discuss the reflection of shock wave reflection of shock wave we will discuss now the reflection of oblique shock what is that reflection of oblique shock wave oblique shock wave reflection of oblique shock wave what is meant by reflection of oblique shock wave. Now, it is very important that how the shock wave is reflected. Now, let us consider a case like this physically first of all that there is an wedge there is a wedge there is a wedge being. So, there will be a shock flow is approaching with some v 1 m 1 v 1 t 1. Now, here if I show the velocity so it will go like this and this will be the v 2 turning this is the half angle here also what happens the flow will turn like this. Now, if there is a plate here flat plate for example. So, first I consider reflection by a flat plate simple case reflection by a flat plate then what will happen physically you see that this thing will ultimately become parallel to the plate after what the streamline will become parallel to the plate. This means that the shock comes here this is the shock wave shock wave strikes the plate and is being reflected such a way that this when comes to this reflected shock wave. So, this is the deflection for example, this value is delta. So, this is deflected this is again this is this direction deflected in the same value delta to make it parallel. That means initially this was the flow direction which was parallel to the plate for example, it is deflected by delta by the original shock wave this is original shock wave well and this is the reflected shock wave reflected shock reflected shock wave well. So, therefore, you see the original shock wave deviates this one by a angle delta that is the turning angle of this that is the angle imposed by this wedge shape body. So, then what I mean this side is a plate then again the flow stream will get reflected by the reflected shock wave same angle of deflection to make it parallel. This is a typical picture very simple case that how does a oblique shock wave gets reflected by the flat plate. Let us solve a problem based on that then only this things I think reflection of shock wave will be better clear from a problem. Let us see this problem example 2 well air flowing at m is equal to 2.5 mack 2.5 with a pressure of 60 kilo Pascal and a temperature of minus 20 degree Celsius passes over a wedge which turns the flow through an angle of the folding. Simple case that means there is a oblique shock takes place attached oblique shock to the wedge this is again told in the problem that leading to the generation of oblique shock waves this is true. One of the oblique shock waves as I have just described the same physics is being repeated in the problem with numerical data impinges on a flat wall which is parallel to the flow upstream of the wedge and is reflected from it find the pressure and velocity behind reflected shock wave that means this is the problem find the pressure and that means precisely this is the problem let this is 1 let this is 2 let this is 3. So, this we have to find out what is p 3 now what is given p 1 m 1 is 2.5 well p 1 is 60 kilo Pascal and t 1 is minus 20 degree Celsius these are the given data and delta is delta in both the cases is 4 degree it becomes again parallel to its original direction which is parallel to the wall this is the wall plane wall problem is very very simple first we find out the condition at 2 now m 1 is to first duty is to find out beta that means at delta is equal to 4 degree and m 1 is 2.5 I find beta the weak shock solution beta the lower value of beta and from the figure and this becomes equal to 26.6 degree that means I just search for this m 1 is 2.5 I go somewhere here that is the m 1 is 2 this is 3 and with the beta 4 degree we find out this find out this m 1 is 2.5 somewhere here it is 30. So, it may be 22.6 however I am not doing it here but this was found as this is the correct value. So, whenever I know beta so this side m 1 is m 1 sin beta m n 1 is m 1 sin beta whose value is 1.12 now I can read from normal shock table that this 1.12 what is the value of p 2 I can find out p 2 by p 1 p 2 by p 1 and that p 2 by p 1 that is found out is equal to 1.336. And t 2 by t 1 is found out this is all from normal shock table because now it is equivalent to a normal shock I am reading m 1 sin beta that 1.12 as m 1 in the normal shock table to find out p 2 by p 1 and t 2 by t 1. Therefore, I can find out p 2 and t 2 here because p 1 and t 1 is are given. Now, what will happen I have to find out the m 2 here how to find out m 2 here corresponding to this m n 1 I know an m n 2 from the table that is the m 2 from the shock table and that value is 0.897 that means this is the value of m 2 in the shock table that means the downstream Mach number corresponding to an upstream Mach number of 1.12. So, corresponding to an upstream Mach number of 1.12 I find the downstream Mach number m n 2 which downstream Mach number m 2 which is our m n 2 which is actually m 2 sin beta minus delta here beta I know 26.6 this is this value I know the 26.6 this is this is the 26.6 this is 26.6 this is 26.6 this is the beta we know. So, delta is 4 degree. So, we can find out m 2. So, what is m 2 m 2 we find out as 0.897 I write it sin of 26.6 minus 4 degree and that become equal to 2.33. Now, is another problem that second reflected shock wave that m 2 is 2 that is approaching with a m 2 2.334 which is being deflected this way by delta. So, therefore, now again with delta is equal to 4 degree and this m 2 as m 1 that means approach Mach. That means this m 2 that is here it will act as m 1 for this table we find out what is beta. That means for this table when we will use to find out this table this table this Mach number will be this one this 2.334. So, for this the shock wave angle will be 28.5 degree that means for the reflected shock this shock wave angle is different. So, shock wave angle is 28.5 degree this is now this is 28.5 degree now turned again by delta is 4 degree do with this 28.5 degree I decode again m n 2 this m n 2 is what this is with this approach m 2.334 which is our m 2 which I am using as if m 1 in this shock table 2.334 approach or to find out the value of beta with the delta m 1 that figure delta beta for different m 1 that figure. So, m n 2 is 2.334 into sin of this 0.5 I think you will understand this thing clearly. That means this becomes the sin component the normal component of the Mach number though Mach number is a though you should not use that what component Mach number is a scalar quantity it is the ratio of the velocity. But this is the way you can understand that the sin component of the Mach number which acts as the normal shock Mach number now with this m n 2 this becomes equals to I tell you sin this is sin 28.5 and this becomes equal to 1.111. Something like that you see 1.1 as I found find the solution now against this Mach number in the shock table. Now, this is the approach Mach number for the reflected shock I can find out the pressure ratio that means I can find out the p 3 by p 2 and this p 3 by p 2 I can find out 1.297 and I can find out t 3 by t 2 which will be read as t 2 by t 1 they are in the shock table. That means I will use this approach Mach number in the shock table as m 1 and find out p 3 by p t 3 by t 2. Earlier we found out p 2 t 2 because we knew that p 2 by p 1 and t 2 by t 1. So, therefore, we find out p 2 t 2 multiplying with p 1 and t 1 then p 3 by p 2 and t 3 by t 2 is known. So, we can find out p 3 t 3. So, final result is that p 3 is equal to 104.0 kilo Pascal and t 3 is 296 k. Now, velocity has to be found out. So, find the pressure and velocity behind the reflected shock how to find out velocity velocity is very simple you find out a 3 at root over gamma r t 3 well root over gamma r t 3 and when you get the velocity a 3 you get the v 3 as a 3 into m 3 what is m 3 m 3 is this one m n 2 you get here and you find out the corresponding value of m 3. That means you find out here a value of m n 3 from this with this value of the approach Mach number and that m n 3 value equals to I tell you 0.9108 0.9108 0.9108 then you find out m 3 decoded that is m n 3 is m 3 divided by that is m n 3 m 3 into sin beta minus delta. So, therefore, sin beta this beta is the beta for the reflected shock wave that means this beta 28.5 delta remains the same in both the cases. So, therefore, you get a value of m 3 is 2.17 fine. So, the m 3 if you put a 3 if you put you get a value of v 3 same as m n 3 0.9108 0.9108 0.9108 0.9108 0.9108 0.9108 0.9108 0.9108 0.9108 0.9108 0.9108 0.9108 0.9108 0.9108 0.749 meter per second. So, therefore, we find out the pressure temperature and everything. So, you see that a Mach of 2.5 is ultimately reduced to Mach of 2.17 because we have chosen always the weak solutions always the weak solutions. I am also not sure which one will prevail, but we can have the strong solution is just only a change in the value and we will get the some we will follow the same procedure, but finally experiments we will tell which one is correct. So, this is the way we have to solve the problem. So, therefore, it is clear now again I am telling before closing the lecture on oblique shock that the oblique shock has to be treated always as a normal shock by considering the fact that the this the velocity component parallel to the oblique shock does not change. So, therefore, the normal component of the velocity acts as the same thing that is the that is the velocity of approach for a normal shock. So, we consider this Mach number M 1 sin beta if beta is the shock wave angle and M 2 sin beta minus delta if delta is the deflection angle as the Mach number before and after the shock and we can treat this as a normal shock that is the main thing and delta the turning angle is imposed by the problem geometry for which one has to find out beta depending upon the inlet Mach number M 1. This reduced Mach number is for using the normal shock table that means to convert the problem to a normal shock problem. So, this is the total thing for the oblique shock. So, therefore, today I will stop the oblique shock discussion and now I will start the another chapter I do not know whether the time is not or is there or not, but rather let me start at least the now the next question is that so far we have discussed about a discontinuity in the flow field where the pressure changes in a way that pressure is decreased and the supersonic flow is changed to either a supersonic flow or a subsonic flow in the form of deceleration either a supersonic flow or subsonic flow in a form of deceleration for a normal shock it is it has to be subsonic for an oblique shock it may be subsonic it may be supersonic, but it has to be decelerated where pressure temperature will increase. This type of wave is known as compression wave where the velocity is decreased and the pressure is increased and this type of wave may be stationary with respect to a static from a frame of reference or with respect to a coordinate frame of reference you can say and this can be moving also and in that case moving with a constant velocity we can attach the coordinate frame with the shock wave and we can analyze, but there are certain cases where this is not this compression wave does not occur. For example, I tell you just consider that so far we have considering a O H shape for example, a O H shape where the the direction is changed like that. Let us consider a we have considered a shape like this ok. So, this is the case where we have seen that there is a this we have not considered there will be a shock wave like this oblique shock wave where this will be there will be change like this the deflection ok. So, P 2 this is P 1 this is V 2 this is V 2 this is V 1 and it is true that V 2 is less than V 1 and obviously, P 2 is this has been proved P 2 is T 1 here also the oblique shock wave is there. So, that sorry that this difference and V 2 V 1 and V 2 is less than V 1 this is the wage and P 2 is greater than P 1, but what happens if a supersonic flow approaches a convex corner it may be smooth it may be sharp what happens. That means, this what happens what will happen that if this is the thing what will happen will there be a shock wave and this will be velocity will be turned like this and V 2 will be less than V 1 or V 2 will be greater than V 1 which one will prevail which one will prevail tell me. So, this is here we can show from the geometry that V 2 will be greater than V 1. Another example I am telling you that if there is a pipe this pressure is atmospheric pressure this pressure is P which is much greater than atmospheric pressure there is a valve or diaphragm whatever you take it is suddenly raptured it is made open made open the fluid will suddenly come out. Then what will happen this side the pressure will be released and velocity will suddenly slowly slowed in a compressible flow depending upon the extent of compressibility the velocity in the upstream will slowly be reduced and pressure in the upstream will slowly be reduced. So, what will happen so this means an expansion wave will move in this direction here also this will be a static expansion wave means this V 2 will increase whereas, the pressure will fall here also the fluid will start flowing that means there will be a that means what will happen in the upstream velocity will be generated because the flow has started suddenly open the discharge will take in an incompressible flow the same discharge will be instantaneously felt here, but in a compressible flow it is not felt here or the pressure felt here is P a then immediately the pressure will not be felt here as P a. So, therefore this will take time that means there will be a propagation of a wave which is an expansion wave why which makes the pressure low as it passes through and makes the velocity high because it was 0 similar thing happens this is a stationary wave where this V 2 will be higher than V 1 I will explain this in this case if it is a convex wedge or a convex corner rather sharp or smooth corner this type of oblique shock this is oblique shock will not occur this is oblique shock. So, this is oblique shock what happens this will be expansion here we will get expansion wave which I will discuss in the next class I think time is up. So, today we will finish it here ok thank you.