 Meaning 14 and welcome all to this session of discourse. So, far we are discussing the show normal shop. And in all those discussions we have made this actually at locations. While the fluid flowing and across the shop there were these年的 In the fluid properties that we have seen. Now, there are situations where the shock is moving sometimes in a flow relative to a flow, flow may be rest at rest, flow also may be moving which are observed in certain practical cases like in air compressors which we discussed in turbo machines, sometimes in the exhaust and inlet manifold of internal combustion engines which may occur due to explosion, a shock wave propagates through a medium which may occur in case of fluid flowing in a pipeline, we just close or open the valve at the downstream, so a shock wave is generated which is propagated or moving to the in the fluid medium. In those cases it is easy to analyze the mathematical relations or to analyze the problem with respect to a coordinate system attached to the shock moving with uniform velocity, then the entire thing appears to be a stationary shock wave across which there is a change of fluid properties, but how these analysis are being made and results are being obtained, we have to see. Most important thing here in this context is to find out the velocity imposed by the motion of the shock wave to the fluid medium, so to understand this let us consider a case like this, let us consider a case like this rather I write here moving normal shock, moving normal shock, moving normal shock wave, shock wave. Let us consider a case like this where in a duct a shock wave, this is the shock, this is the shock wave, this is the shock wave, let us consider a situation like this that a shock wave is moving in a duct with a velocity of u s constant velocity of u s and the gas behind the shock that means downstream part of the shock it is at rest there is no velocity v is 0 and pressure is p 1 and temperature is t 1, whereas this shock has created a velocity in the fluid through which the shock has passed, let this velocity is create a velocity v, let us consider a velocity v, let us consider a velocity v which is being and it has created a pressure p 2 and temperature t 2, this is the velocity of flow v that means due to the shock passing over this fluid a velocity is imposed or induced by the shock. Now, this problem if you analyze in a way that this is a picture with respect to a coordinate frame attached to the wall that means a coordinate frame static frame of reference which is at rest. Now, if we attach the coordinate frame on the shock that means if we see this thing relative to the shock it is just like a stationary shock wave normal shock wave that means shock is at rest here that means then it will look like that the fluid velocity u s and a pressure p 1 and temperature t 1 approaches the shock and while it goes this side it is velocity is reduced u s minus v usually this u s is higher than v and therefore there is a deceleration in the same direction and this is u s minus v and the pressure is p 2 and t 2. So, therefore with respect to the coordinate axis attached to the shock wave this is the picture. Now, if this you make an analogy with the stationary shock then what you see you see that in the stationary shock the approaching velocity is given as v 1 which equals to u s and this is v 2 which equals to u s minus v then p 1 p 2 that means section 1 and section 2 then v 2 is u s minus v. So, now if this is the thing now if we define the Mach number at the inlet that is upstream of the shock then what is the Mach number now here I can write the Mach number m 1 as u s by a 1 where a 1 is the acoustic speed sorry a 1 is root over gamma r t 1 well a 1 is equal to root over gamma r this is known as this m 1 is u s again I am writing a 1 is known as m s this is known as this is just a definition shock Mach number shock Mach number a shock Mach number m s. Now, what is m 2 now that means with respect to a stationary wave. So, this is the picture so m 2 will be the velocity after the shock wave that means u s minus v by a 2 where a 2 is what a 2 is root over gamma r t 2 clear this can be written as u s by a 2 which can be written as a 1 this is the style of writing a 1 by a 2 minus v by a 2. So, this v is the v generated due to the passing of the shock wave, but in a steady condition with reference to a coordinate frame attached to the shock wave that means if you make the shock wave stationary by imposing a velocity u s in the opposite direction to the entire system this is the situation. So, I am writing in that case the m 2 is the velocity that means the Mach number after the shock that is the stream velocity u s minus v. So, this become this one that means this is equal to m s into a 1 by a 2 minus this is v by a 2 minus v by a 2 minus a 2 is written as m 2 dash that means m 2 dash is nothing, but v by a 2 that means this v by a 2 is written as m 2 dash this is an important parameter. So, therefore, this relationship you have to first of all remember or you have to write that relations that means it happens that a fluid with a Mach number m 1 given by u s by a 1 and with a velocity u s and p 1 t 1 is passing through a stationary shock wave and its velocity is reduced that is Mach number rather is reduced to m 2 which is defined as u s minus v by a 2. Now, this quantities are with respect to this problem this is an unsteady problem where the shock is moving and these velocities are defined with respect to the coordinate frame static coordinate frame which is attached to the wall coordinate frame attached to the wall. Now, if we accept this now next part is that if we recall the analysis we if we now recall the shock relationships just I am writing this in earlier classes in few earlier classes we have discussed this thing that the suffix 2 is always the so now I am writing the shock relations normal shock relation normal shock just recapitulation normal shock relations recapitulation of normal shock relations this equal to this p 2 by p 1 that means the suffix 2 as you know is a usual symbol is after the shock and this is before the shock usually in shock more useful relationships are the ratios of pressure temperature density in terms of the inlet Mach number inlet Mach number is the input parameter. So, flow is usually specified by inlet Mach number so I am recollecting this formula this formulas that we already already deduce gamma plus 1 similarly rho 2 by rho 1 has an expression which is gamma plus 1 m 1 square divided by 2 plus gamma minus 1 m 1 square similarly we can write t 2 by t 1 which was all deduced earlier this is this 2 plus gamma minus 1 m 1 square if you remember this relationship 2 gamma m 1 square minus gamma minus 1 and divided by gamma plus 1 whole square m 1 square now this t 2 by t 1 can be written also equals to what is t 2 by t 1 a 2 by a 1 whole square why because the a 2 at the section 2 is root over gamma r t 2 and a 1 is root over gamma r t 1. So, ratio of temperature is the ratio of the acoustic speeds at that state corresponding states a 2 by a 1 square another very important relation is there that is the Mach number after the shock which is the subsonic Mach number gamma minus 1 m 1 square plus 2 2 gamma m 1 square minus gamma minus 1. So, at present I recall only this form this relationships again I tell what are those relationships this is the change in the fluid property in terms of the ratio pressure ratio the temperature t 2 by t 1 rho 2 by rho 1 the density and the Mach number after the shock in terms of the initial Mach number. Now, this relationship are exploited to find out a chart where against a Mach number 1 that is m 1 against m 1 against a Mach number m 1 we can found we can find all these properties. Another thing you have to remember with this or the recapitulation which has been already discussed earlier from the consideration of entropy change as a corollary of the second law we always tell that m 1 has to be greater than 1 for a shock to occur and m 2 will always be less than 1. That means a shock always decelerates the fluid approaching the shock from a supersonic one to a subsonic one after the shock. So, this if you recapitulate or you remember now here what happens now this problem this is similar to a stationary shock where m 1 is this one m s and m 2 is this one m s a 1 by a 2 minus m 2 dash or you can write u s minus v by a 2. So, this is our m m 2 now what you do if you write the same this replace this you get this. That means instead of m 1 you replace m s that means in this case for this moving shock wave I can found p 2 by I can find p 2 by p 1 just again I am writing just replacing this in terms of m s that means rho 2 by rho 1 I am writing little fast because it is just for your taking note m 1 is because m 1 in this case is m s u s by a 1 m s square. So, according to the present nomenclature I write this one and t 2 by t 1 will be t 2 by t 1 will be again same that m 1 is replaced by m s m 1 is replaced by m s that means 2 plus gamma minus 1 m s square into 2 gamma m s square minus gamma minus 1. So, this is also equals to a 2 by a 1. So, this a 2 a 1 related to t 2 t 1 this t 2 t 1 p 2 p 1 these are scalar quantity they are not dependent on the coordinate transformations. Now, the m 2 similarly has to be replaced if you write the substitute equation for this then what will be the equation this equation will be instead of m 2 this equation we write here m 2 is m s that in our earlier thing that m 2 is m s a 1 a 2 minus m 2 dash square. So, therefore, m 2 square will be there for m s a 1 a 2 minus m 2 dash this is the square and that equal to same thing just we are replacing the corresponding m 1 m 2 where 1 2 is the upstream and downstream section of the m 1 square minus gamma minus 1. So, this is our m 2 at the present moment m s a 1 by a 2 minus m 2 dash now what happens if we take this equation and if we use the a 1 by a 2 then we get a very important relationship of m 2 dash let us do that. Now, let us again write this equation let us write this equation again now let this equation is written you just remember this equation this equation is written this equation again I am writing m s a 1 by a 2 minus m 2 dash I take a square root and I write like this I think you can see that thing I write that gamma minus 1 m s square plus 2 to the power 0.5 and here 2 gamma m s square minus gamma minus 1 to the power 0.5. Now, a 1 by a 2 is what a 1 by a 2 is the under root of this thing. So, you just see that this is the from your note that as you have taken I am writing then m 2 dash is equal to m s into a 1 by a 2 what is a 1 by a 2 is under root of that. That means, a 2 will be m s this is a 2 by a 1 that means a 1 by a 2 is the under root of reciprocal of that that means denominator will be 2 plus gamma minus 1 m 1 square to the power 0.5 into 2 gamma m 1 square minus gamma minus 1 to the power 0.5 that means numerator will be denominator it is a 2 by a 1 I am writing a 1 by a 2 and under root because I am this is square. Now, the denominator will be under root gamma plus 1 whole square m s square that means this gamma plus 1 whole square that means gamma plus 1 into m s and denominator it will be m s very good denominator it is m s very good denominator it will be m s very correct so denominator it will be m s all right. So, m 2 dash is m s into this minus if you do that you will get gamma minus 1 m s square plus 2 to the power 0 plus 1 whole square 0.5 divided by 2 gamma m s square minus gamma minus 1 to the power 0.5 you can see it very correctly then this becomes equal to what then this becomes is equal to m s square gamma plus 1 so this will be the denominator 2 plus gamma minus 1 m s square to the power 0.5 let me write this 2 gamma m s square minus gamma minus 1 to the power 0.5 then minus so this 1 is 2 gamma that means this 1 2 plus gamma minus 1 m square 2 plus that means this will be simply 2 plus gamma minus 1 m s square that means the same thing 0.5 0.5 so if you now clear this thing then the denominator will be what denominator will be this big thing 2 plus gamma minus 1 m s square it is little tedious I know but you just follow that whether I am doing any mistake or not m square minus gamma minus 1 to the power 0.5 and now m s square gamma plus 1 minus 2 gamma minus 1 so gamma m square gamma m square is cancelled so this is m square and then gamma now if you write this gamma here I make a calculation gamma m s square plus m s square minus what is that 2 plus gamma minus minus gamma m square plus m square that means this cancels that means 2 into m square minus 1 this become 2 into m square minus 1 so this is 1 very important deductions that is here I can write m 2 dash I think you can see equals to which is very important m 2 dash why I am writing it is v by a 2 it is not m 2 m 2 is this 1 m s a 1 a 2 minus m 2 dash this is m 2 but I am finding out an expression of m 2 dash in terms of m s now you see what interesting results we have achieved now you see here what it is given that it is given not the downstream Mach number that means Mach number after the shock with respect to the shock the coordinate respect to the shock but it is given the ratio of the velocity induced because of the shock in actual case divided by a 2 m 2 dash in terms of the inlet or shock Mach number here m s you remember is u s by a 1 that is is equal to actual m 1 in this case now is a very interesting result it gives now what interesting result that if this if we increase the Mach number let us consider the inlet Mach number is very very high inlet Mach number is very very high that means the flow approaches with a very high supersonic velocity and as you know the strength of a shock as I told you earlier the is expressed by the pressure ratio that means the ratio of pressure after the shock to that before the shock and it is a direct function of Mach number I tell you is a direct function of inlet Mach number more is the Mach number more is this ratio that means the strength of the shock is usually expressed by this ratio which is a function of inlet Mach number or approaching Mach number more is the Mach number more is the strength of shock so therefore a Mach number approach a shock a fluid approaching with high Mach number to a shock wave is told as the shock is of very high strength so therefore when the shock Mach number is very high we usually call that this is a shock of high strength now whatever high strength shock comes there is a limiting value for m 2 dash physically now look it mathematically is there any limiting value when m s tends to 0 if I ask this question that when m s tends to sorry infinity what is the value of what is the limit rather m s tends to infinity what is the limit that limit I will now show you now m s tends to infinity m 2 dash can be written as so this type of thing as you know elementary mathematics at school level when m s tends to infinity let it expressed in terms of 1 by m s that means both numerator and denominator should be divided by m s square this is the standard technique see that whether the limit is there or not so m s square means 1 m s I can take here 1 m s I can take here and when it goes within the bracket again m square that means this will be 2 by m square that means this gives directly an indication that there is a finite limiting value the term is becoming free of m s m s by this m s we are dividing numerator and denominator by m s square this m s goes inside so therefore m s square here also 2 gamma minus gamma minus 1 so beautiful that means this give us an indication what we want that there is a finite limiting value so when m s tends to infinity it is 0 it is 0 and it is 0 so the limiting value there is a limiting value is root over 2 gamma into gamma minus 1 this is gamma minus 1 2 and 2 under root and this is 2 so therefore 1 2 under root will go up so therefore you see the limiting value that I was telling that limit m 2 dash or equal to m s square tends to infinity limit v by a 2 as m s that means the shock mach number tends to infinity is not infinity it is square root of 2 by gamma into gamma minus 1 this is a beautiful result and you see that for air for air if we take gamma 1.4 this m 2 dash is 1.8 that means you can say that in air whatever strength may be of the shock even a infinite strength shock very high strength shock can only induce a finite velocity which will be always less than this corresponding mach number 1.89 the actual velocity will depend upon the temperature we have to multiply this with a 2 which is root over gamma r t 2 that means velocity is usually specified in compressible flow by the mach number so you can tell for example in air whatever strong mach number may be at the upstream of a moving shock this can or whatever rather this way you tell whatever strong shock may be a moving shock whatever strong the shock may be whatever velocity the shock may be moving that the m s u s by a it cannot induce a flow whose mach number will be more than 1.89 in case of air so this is the limiting mach number in case of air now next is next we will rather now after this I think we will solve some problem so that your all these things will be made more clear by solving the problem before I go to a this reflection of the shock let us solve some problems. Now problem one example one a this is the first problem example say shock wave across which just you see this thing problem which the pressure ratio is 1.25 this is not a this is a moving shock a shock wave across which the pressure 1.25 is moving yes it is written is moving into steel air at a pressure of 200 kilo Pascal s and a temperature of 15 degree Celsius this is the steel air find the velocity pressure and temperature of air behind the shock wave that means it is simply the application of that how to do it now let us again solve the problem the shock is moving with a velocity is moving into steel air and we do not know its velocity let u s similarly that u s and this p 1 t 1 the same problem and then it induces velocity v and p and p 2 p 2 p 2 so with respect to the shock the problem is with respect to the shock the problem becomes steady with a velocity u s p 1 t 1 and it goes with the velocity u s minus v and p 2 t 2 what are given in the problem p 1 p 1 is not given rather p 2 by p 1 is 1.25 what are the given t 1 is given 15 degree Celsius p 1 is given 100 kilo Pascal what we have to find out velocity pressure air behind the shock that means we have to find out v we have to find out pressure p 2 and temperature t the first one is the school level thing p 2 because the pressure ratio is given 1.25 I know the pressure p 1 so p 2 will be 100 into 1.25 125 kilo Pascal straight way but we do not know t 2 how to calculate now I have told you earlier that in shock all those relationship we have derived that means for example if p 2 by p 1 is known I can find out m 1 because p 2 by p 1 is expressed in terms of m 1 again I tell you that just also at the beginning of this class I told you that all these relationship which was derived earlier for example this important relationship here one can find out algebraically p 2 by p 1 I know I can find out m 1. So if I know m 1 I can find out rho 2 by rho 1 t 2 by t 1 over this algebraic calculation sometimes are tedious. So therefore a table is drawn based on this calculation as usually drawn for different cases that table as I now tell you earlier also I told which is known as normal shock tables. So this is there just you see how to see a normal shock tables a normal now I not will do here I will rather do here a normal shock table we just how does it look like you can see in any book normal shock table just I show you probably I told you earlier also normal shock table looks like that in the left hand column these are the different column m 1 the corresponding nomenclature you know that is the mach number approaching the shock with respect to a stationary shock this is the after shock mach number then this is the ratio of pressure 2 is the section after the shock and this is all the ratios of the static pressure temperature the density and there is a ratio of stagnation pressure also because of the losses shock is an across the shock the process is irreversible. So therefore p o 2 by p o 1 now these there are number of mach numbers different values starts from 1.00 why because shock does not occur when the approach mach number is subsonic. So therefore it goes some for example 5.00 so corresponding number all these things are there. So then if I know the p 2 by p 1 now you have to see that what are the quantities given in this problem we know p 2 by p 1 we know p 2 by p 1 is 1.25. So if we know p 2 by p 1 I can find out everything m 2 t 2 by t 1 that means any one of these is known other things are known through the equations. So therefore it is not that always m 1 has to be known anything has to be known we can find out from the chart. So from this chart if you read with p 1 p 2 by p 1 1.25 you will get a mach number of m 1 which is which from normal shock table you will always get m 1 and m 2 m 1 means inlet mach number m 2 means outlet that means the downstream after the shock mach number 9 and t 2 by t 1 you get 1.0662. Now when you get t 2 by t 1 straight t 2 is t 1 is given t 1 is 15 that means 288 into 1.0662 so you get t 2 so t 2 is 307 in this case. Now in your case m 1 is what m 1 is u s by a 1 now what is a 1 a 1 is root over gamma r t 1. So therefore u s is the mach number I have got from the table into root over gamma that is 1.4 into r is 287 into the t 1 t 1 is 15 degree means 288. So therefore you straight forward get the value of u s if you know m 1 already you know a 1 root over gamma r t 1 and you get the value of u s. Now you have to find out the velocity behind the shock that means this velocity it is very simple to do this what you have to do you have to write the expression for m 2 in your case in your case m 2 is what is m 2 u s minus v which you have to find out by a 2. So therefore you can write v is equal to a 2 a 2 u s minus m 2 a 2 u s is always I know that this is found 1.102 into root over 1.4 into 287 is the value of r into 288 minus m 2 which I have found out 0.9103 into again a 2 a 2 is root over 1.4 into sorry 287 into root over 1 into t 2 t 2 I have already calculated because I know t 2 by t 1 from the table. So t 2 is calculated 307. So most important parameter is this we want to know this this comes out to be 55.2 as far as this calculation I know the answer this is the answer. So this is a straight use of this formula which we derived the analysis which we have made for a moving shock. Now another interesting problem you see this is a relatively more interesting problem this problem what does it say what does it say a normal shock wave across which the pressure ratio is 1.17 moves down a duct into steel air at a pressure of this is almost the same and a temperature of 30 degrees rest part is little interesting find the temperature pressure and velocity of air behind the shock wave this part is routine. That means you can say that said this is the total repetition of the earlier problem with probably difference in the data. Now next one this shock wave passes over a small cylinder circular cylinder it is actually it will be here circular cylinder as shown in figure that I will show you the figure assume that the shock is unaffected by this small cylinder fine pressure acting at this stagnation point on the cylinder after the shock has passed over it. Now this is a problem here now this problem if you solve then how you what is the problem problem is this one that this is the shock shock at this is a small cylinder. So, shock has started from somewhere else in the upstream it has passed different sections at any instant as the problem tells when the shock has passed over the cylinder and shock remains unaffected what is the condition. That means in the entire upstream part of the shock through which the shock has already passed the fluid this has attained some induced velocity v now let us find this induced velocity v. So, therefore our case is similar to this then our case is similar to this that let this is the cylinder our case is similar to this. So, u s same thing u s minus v this is 1 and this is 2 and let this is p 2 t 2. So, p 1 t 1. So, I can find everything, but first of all I have to see with respect to this stationary shock what is the thing in pressure ratio same thing that means I am having pressure ratio that means I can immediately go to this shock table normal shock table and see against this pressure ratio what is my value m 1 m 2. So, this pressure against this pressure ratio the m 1 as found from the shock table is 1.07 m 2 as found from the shock is 0.936 and t 2 by t 1 as found is 1.046. So, therefore since t 2 by t 1 is known I can immediately calculate the temperature value temperature behind the shock that means this temperature t 2 I can find out t 1 is what t 1 is 30 degree that means t 2 is 1.046 into 273 plus 30 that means 303. So, I can find out this is this equals to 316.9. So, I know this and p 2 I can find out how p 2 is equal to this is the school level thing 1.17 it is already given 1.17. So, 1.17 into 105 kilo Pascal. So, I can find out and this becomes is equal to 122.9 kilo Pascal. Now, what I have to find out next assume that the shock is unaffected before that I have to find out the velocity of air behind the shock this is again the same thing that m 2 I know that means m 2 is 0.936 is equal to what u s minus v divided by a 2. Now, a 2 is what a 2 is equal to root over gamma r t 2 that means 1.4 gamma r 287 into t 2 because t 2 is 316.9. So, I get the value of a 2 I will just substitute here the value of a 2 m 2 that means I can write better v is equal to u s minus m 2 a 2 that means u s what is u s u s is again m 1 is what is m 1 here m 1 is u s by a 1 and a 1 is root over gamma r t. So, u s is m 1 times a 1 root over gamma r t t is 3 0 3 minus m 2 m 2 is 0.936 times root over gamma r and t 2 that means 3 16.9. So, that is very simple 3 16.9 you understand 3 16.9. So, therefore, I get the value of v that this v becomes equals to some 33 this becomes equal to v becomes equal to 39.38 meter per second I think it is absolutely all right. So, it is just the repetition of the earlier problem now the question is this that we have to find out the stagnation pressure now what happens if you see this picture now this cylinder actually here this is the model physically translated, but actual problem is this. This is the problem as we look with reference to a coordinate frame attached to the moving shock and flow becomes steady and this becomes stationary. So, actually this is the problem this v we are getting which is 39.8 that means now it is a problem that a cylinder circular cylinder is exposed here another thing is told stagnation point assuming that the shock is unaffected by this small cylinder find pressure acting on this stagnation point on this cylinder after the shock has passed over it and there is another thing which has to be told that consider this flow to be isentropic consider the flow the pressure acting at this stagnation point on this cylinder assuming that the flow is isentropic. Now, if we assume the flow to be isentropic then what happens there is an isentropic flow past a cylinder whose velocity is 39.38 meter per second. So, we can find out the stagnation pressure before that I have to tell you that if you recall the if you recall this isentropic flow relations isentropic table that isentropic flow relations are unnecessary without going to all algebraic calculations through the algebraic equations of isentropic flow relations I rather isentropic flow isentropic flow table that is normal shock table it is given in a tabular form similar to normal shock table that one column the extreme left column is m then these are the stagnation because here most important part of calculation is the ratio of stagnation temperature to static temperature stagnation to static all stagnation property respect of density to the corresponding static is the function of the mach number of flow. So, therefore, if we know the mach number of flow at that particular location we can find out all this thing similarly a 0 by a what is that a acoustic speed at the stagnation condition to the local acoustic speed a by a star probably you recall this things a star is the area where the sonic velocity is rich that is the minimum area known as throat area corresponding to m 1 and this a is the area local area at which we are concerned with and another parameter is theta I told probably earlier that does not come into picture here that will come when we will discuss the expansion wave this is the angle related to the expansion wave this is not coming at that there also the problems are treated as an isentropic flow that is why in the isentropic flow table itself this theta is given this is not required now. So, now if we know the mach number at a particular location in an isentropic flow this here mach number start 0.01 0.04 like that very small because all subsonic supersonic all levels are there. So, there will be somewhere 1 when the value of a by a star will be 1 this things probably I have discussed earlier. So, now if we know the local mach number I can find out the p 0 by p t 0 by t I can read that everything. So, here I have to find out the local mach number now how to find out because I know now the actual problem the cylinder is exposed to this uniform velocity 39.3. So, local mach number is m for example here is 39.38 by the local acoustic speed which is root over 1.4 gamma into r into this temperature that means this is 316 that means it is nothing, but a 2 it is nothing, but in our this problem this is a 2. So, you find out this m and this m if you calculate this m will become equal to I tell you the value which is already calculated by me this m has got a value this m if you find out this m is I tell you just wait I will tell you the value of m this is a problem where the m 2. So, therefore, we take a value of m this m is equal to this is a v 2 and a pressure the mach number is 0.11 just I see the calculations. So, that you can check 0 point it has been calculated by me earlier. So, 0 point see that thing. So, therefore, in the table if you see this mach number you can find out the value of p 0 by p and that p 0 by p if you see from the table this also I tell you for this problem is this. So, therefore, you can very well find out p 0 is 1.0085 into the static pressure there which is p 2 122.9 because the condition prevailing is this p 2. So, this becomes is equal to 123.9. Let me see the value 123.9 as I have calculated already kilo Pascal. So, this is the value. So, it is clear that you can find out the value of the mach number and you can find out the value of the stagnation point pressure. So, this is one very interesting problem. Now, I will tell you something regarding the reflection of a shock wave regarding the reflection of a shock wave. What is reflection of a shock wave? So, what is reflection of a shock wave? Let me tell you what is reflection of a shock wave. This happens when the shock wave is moving. So, moving shock wave analysis we have done reflection of a shock wave. Now, reflection of a shock wave if we have to understand then just how does it happen? I tell you the physical picture now if there is a cylinder we have so far considered that it is moving with u s and it is creating a velocity v and a pressure p 2 t 2 and this side the pressure is p 1 and initial velocity v is 0 and temperature t 1 that was our thing that it is moving in this direction. So, with respect to these things are happening. Now, this is the actual problem physical problem as we are discussing. Now, what happens if this thing is closed at the end? That means there is a closed end or the pipe was open suddenly the valve is closed. That means the shock wave is moving and the pipe is closed at the end or is being closed by the closure of a valve at the end. So, what will happen? Physically you try to understand what will happen? This velocity will be induced so long the shock moves. Now, try to understand what happens in reality now when the shock comes here the velocity will be imposed v, but after that what happens this is the closed end. So, at this solid end the velocity will be 0 there has cannot be any velocity the velocity cannot penetrate into the solid like that. So, there will be no normal velocity. So, therefore, this velocity will be 0 and slowly in a compressible fluid the entire fluid will attain a velocity 0. For an ideally incompressible fluid this will be instantaneously 0, but for a compressible fluid the velocity will first fall here 0 and it will take some time which may be infinitely small depending upon the compressibility of the fluid that the entire velocity will again be 0. So, this thing is perceived in a way as if the shock wave after reaching there is being reflected back in a way that it creates a 0 velocity here is reflected back in a way that usually the velocity it created in this direction v again it created a velocity v in this direction in the opposite direction. So, that the final velocity is 0 that means the strength of the shock will be such for the reflected way. So, that it can create a 0 velocity while passing through it. So, that is the philosophy of the shock reflection of the shock. Now, let us do that reflection of the shock let us do that the reflection now if reflection of the shock is to be understood let us consider not the initial one the shock is being reflected now. The shock is being reflected now and we can just consider a case that when the shock is reflected and this is the reflection that reflected shock velocity and here let 1 and 2 if we see that 1 sorry this is why 1 and 2 I am telling that this can be expressed as we did earlier in this way that this is coming with us p 1 t 1. So, this 1 and this is 2 this is p 2 t 2 and this is going with us minus v 2. So, now what happened? So, this 2 will now prevail here that is here 2 that is p 2 t 2. So, now this is p 2 t 2 I am sorry now this will be sorry this when you will come here. So, entire thing was then p 2 t 2 that means 2 when it reaches when it reached here the entire fluid was at 2. So, when it is coming here this side is 2 that is p 2 t 2 I think you have understood and this side is 3 that is p 3 t 3 and velocity is 0. So, this is the reflected shock that means the reflected shock if we analyze you will analyze this way. Now with the help of the shock tables we can calculate things that means shock will again be reflected back here. So, this is a typical reflection and under certain circumstances shock will go back again. So, there will be a repeated movement of the shock. So, one such reflection of shock problem we will now solve that how a shock moving ultimately moving in this direction and a particular direction ultimately reaches the closed end and back when it reaches the closed end automatically the velocity becomes 0 this is the reality that means this is conceived as a movement of a reflected shock wave with velocity us are such that this velocity becomes 0 and the condition which is being created is p 3 t 3 again it is changed that means from p 1 t 1 this is p 2 t 2 this is with respect to the shock then the entire thing will become p 2 t 2 here also when it reaches here the entire thing becomes p 2 t 2 pressure p 2 temperature then again it will be changed to p 3 t 3 when the reflected shock will pass through that. So, this is the scenario physical scenario I think that this will be more easily understood if we solve a problem and this time we cannot I cannot continue because the time is up. So, in the next class I will solve a problem in relation to the reflection of a shock wave. Thank you.