 good morning and welcome you all to this session of the course. Now, last class we mentioned about the reflection of a normal shock wave from the closed end of the duct. Now, just now we will be solving a problem to understand how the fluid properties are changed because of this reflected shock wave well. So, let us then concentrate on a problem that the problem is like this a normal shock wave across which the pressure ratio is 1.45 pressure ratio is given normal shock wave moves down a duct into steel air at 100 kilopascals and 20 degree Celsius this is again a repetition of the earlier work problems that a moving pressure wave in steel air. Find the pressure, temperature and velocity field behind the shock wave that is the induced by the shock wave velocity induced pressure and temperature rise this already we did. If the end this is the new part of the duct is closed find the pressure acting on the end of the duct after the shock is reflected from. Now, let us make the physical model like this let us make this way no let us make this physical problem like this this is the first one that this is the duct closed end and let this is the physical problem it is first moving with a velocity u s let this section is 1 this section is 2 here the pressure is p 1 temperature is t 1. Now, this is actually sorry this is actually 1 2 is wrong here I write 1 here because with respect to the shock this will be other way that is why I am writing 1 this is. So, this will be p 2 t 2 that means this is the pressure and temperature after the shock wave has passed through it and induces a velocity v as we did earlier and here this section is 1 I am sorry pressure p 1 t 1. So, this is the initial before reflection of the shock then after that what happens the shock is reflected a shock is reflected shock is reflected let us consider this thin shock thin shock it is reflected with the velocity u s r. Now, what is the picture here that if we translate this in terms of the physical model with respect to the coordinate system attached to the moving shock then what will happen that the similar one it is coming with velocity u s at a pressure p 1 and temperature t 1 that is the static pressure and temperature of the steel air which was initially at rest. So, therefore, with respect to shock as usual u s and then it goes out with a velocity this is the section 2 p 2 t 2 2 1 and this goes out with a velocity u s minus v as we did earlier. Now, when we make this one this is coming then sorry what happens when this reaches the end of the duct the entire duct is at 2 state 2 that is p 2 t 2 all right. So, therefore, this side when it is moving again after the reflection this side is the undisturbed side is 2 p 2 t 2 try to understand and this side let it is p 3 t 3 where the velocity v is made 0 finally, this is the problem that when it is reflected back from this end the velocity is 0. That means, if we make the picture with respect to the shock wave then what will be this picture here the velocity is 0 here it is coming with a velocity u s r, but earlier this velocity when this there is a velocity v 2 here there is a velocity I am sorry v 2 here because when this moves here the situation or the state of the duct is t 2 p 2 and a velocity v v 2 rather we have denoted it by v simply v v. So, therefore, with respect to the shock the velocity will be what that means this is u s r coming in this direction we will give a opposite direction that means this will be the section 2 from where this is the upstream of the shock relative to the shock this is the upstream where the velocity is v plus u s r clear and the downstream velocity which was 0 in actual case when the shock was moving it will be u s r and this is 3 this is p 3 t 3. Now, if we can draw this diagram and can treat this thing properly upstream and downstream of a stationary normal shock then the problem can be done very easily by the application by using the normal shock table. So, therefore, what we do first we find out the pressure temperature now this problem p p 2 by p 1 is given p 2 by p 1 is pressure ratio is 1.45. So, p 2 by p 1 is 1.45 it is given at this p 2 by p 1 1.45 what happened this p 2 by p 1 at 1.45 we have a Mach number m 1 which is equal to 1.1772 and a Mach number from the shock table as I told in the last class 0.8567 and the corresponding t 2 by t 1 equals to 1.1137. Now, what is this Mach number 1 with respect to this now this has to be translated here the Mach number 1 m 1 is u s this is the velocity u s by a 1. So, what is a 1 a 1 is root over gamma r t 1 1.41 into 287 into what is t 1 t 1 is given 20 that means 293. So, this is a 1 what we have to find out first of all we have to find out the pressure very simple pressure p 2 is 1.45 into 100 that is 145 kilo Pascal this is the pressure pressure is 145 kilo Pascal then t 2 as we know t 2 by t 1 is 1.1137 into t 1 is 293 and that becomes equals to that becomes equal to 326.3 k. Now, a 1 is known thus again the same formula that to find out first we find out v this v induced by that how it will be found out it is very simple that m 2 is known m 2 what is m 2 m 2 is just you see here m 2 here m 2 is u s this picture u s minus v divided by a 2 well. So, therefore, we can write now little bit I check it then m 2 is this. So, therefore, v is equal to what v is equal to u s minus m 2 a 2 and what is u s u s we have got it is clearly seen u s now u s is what it is it is seen no u s is what it is seen this is this one m 1 into this one. So, this is m 1 m 1 is 1.1772 this is found from the table into root over this velocity this sounds 1.41 into 287 into 293 this is u s minus m 2 m 2 is 0.8567 into root over gamma r t 2 1.41 into 287 into t 2 t 2 is already we have found out 326.3 well. So, therefore, we get the value of v 2 or v here v the value of v. So, therefore, the value of v as I have calculated I tell you 93.8 meter 93.8 meter per second. So, this is the value of v 2 is the value of v 2 is the value of v 2 is the value of v 2 is the value of v 2 is the value of v 2. Now, you have to see this one this is reflected wave. Now, here if you write here one thing that mac number if you write first of all here what you will write in this case the first of all you write the mac number. Now, the mac number 3 and 2 here actually this two and these two will be different. So, that sometimes well this actually is 3 and this is the stage. Now, I will not use this m 2 now rather this will be confusing already m 2 I have used here. So, therefore, in this situation let us consider this as this coming m m upstream sorry here this is the upstream this is the upstream we called here m upstream and here m downstream m upstream and m downstream m downstream I think I should use a relatively thin one m upstream and m because this 2 3 will be unnecessarily confusing now with respect to this shock wave what is m upstream m upstream is now v plus u s r let us see nothing is known now pressure ratio only this pressure p 2 is known v plus u s r divided by a 2 a 2 is of course, known where a 2 is root over 1.41 gamma into because temperature is already determined 326.3 now what is m downstream m downstream that means here is equal to u s r by a 3 where a 3 is here I write is root over gamma r t root over 1.41 287 into t 3, but t 3 I do not know t 3 I do not know ok let it be like that. So, this can be written as u s by a 2 a 2 is known into a 2 by a 3 u s r by a 2 into a 2 by a 3 now this can be written as u s r by a 2 into root over this is in terms of temperature t 2 by t 3 and this can be little arranged because u s r by a 2 is m upstream minus v by a 2. So, m downstream can be written why I am writing I will tell you after what this is some algebraic rearrangement m upstream this is u s r by a 2 that means m upstream minus v by a 2 this is u s r by a 2 into root over t 2 that means the upstream and downstream mach number is related to each other this is an algebraic rearrangement, but now you see here things are not known, but this can be solved algebraically this is because we know t 2 we do not know t 3 we do not know u s r we know a 2 we know t 2, but at the same time we have the relationships from the shock tables between m downstream and m upstream and also we know the relationship between t 2 and t 3 in terms of the m upstream. So, all these relations together if you write the number of unknowns and the number of equations will be same and you can solve it by a program by writing a program and you can solve it in computer, but I can tell you a very simple way of solving it by iterative technique using the shock normal shock table how is the most simple way of solving it you first make a guess of upstream mach number make a guess any guess. So, as long as soon as you make a guess of upstream mach number try to understand here u s r a 2 a 2 you know. So, you get the value of here v by a 2 you are knowing so since you know this v by a 2 you are knowing so u s r by a 2. So, you know the value of whenever upstream mach number you are assuming you know the value of t 2 by t 3. So, as soon as you know the value of t 2 by t 3 so you can find out t 2 already you know so u s r and you find out m downstream as you can calculate m downstream from here because v value you know again I am telling that you guess a upstream m v you know a 2 you know you find out u s r. So, as soon as you find out u s r what you do now let it be there now as soon as you guess m upstream you get t 2 by t 3 from the shock table. So, you know this one so you find out the m downstream by this equation corresponding to the m upstream there is one m downstream guess m upstream. That means you guess m upstream you get what you get get m down first trial t 2 by t 3. Now, if we know t 2 by t 3 without this m down that you keep it with you then you can find out because v you know a 2 you know because a 2 is this root over gamma r t. So, you can calculate by this formula m downstream you compare with this 2 m downstream. If this m downstream is not equal to the m downstream obtained from the chart for the guess value of m upstream then you modify the m upstream by using the calculated m downstream and repeat the calculation and make a convergence so that everything you can get. That means how you can formulate the problem the solution procedure because you can solve algebraically because if you write all the shock equations relation between m upstream and downstream of a normal shock where all the unknowns will be evaluated from the number of because there will be simultaneous algebraic equations. But simple solution when you have the shock table is by trial and error guess m upstream find out m downstream usr is known. So, you can find out the m upstream so m downstream this way you can find out so when m downstream you are finding out then this m downstream will be compared with the corresponding m downstream from m upstream because guess m upstream will give t 2 by t 3. So, t 2 by t 3 is known v by a square is known for a guess m upstream you get m downstream and again for the guess m upstream you get a m downstream from the shock. So, table so shock table m downstream and this m downstream will compare and then you change the m upstream by the computed m downstream and again repeat the calculation. This clearly understood I hope so this way one can solve a problem of reflection. That means here one thing is made clear I tell you a value if you make this type of iterations then you get a value of ultimately p 3 I write here the value of p 3 equals to 207 find the pressure acting at the end after the shock is reflected from you can find out t 3 also only p 3 is calculated. That means p 3 becomes equal to here 207 kilo Pascal that means when the shock passes here it was at 100 kilo Pascal. So, shock induces a velocity and changes the pressure that pressure equal to 145 kilo Pascal. So, when it is reflected back then again this velocity is created velocity is coming to 0 then again the pressure changes in this side close to the wall and ultimately this pressure will be there throughout when this will come at this end then the pressure will be 207 kilo Pascal. So, you have to set the problem with respect to the stationary with respect to the shock by imposing a velocity in the opposite direction things become steady and a stationary normal shock wave and then you write the equation upstream downstream Mach number and solve the thing with the help of the shock tables or with the algebraic equations. So, with this I will well I will finish this normal shock chapter. Now, I will go to another topic which is oblique shock which is oblique oblique shock oblique what is oblique shock most of the cases the shock which we get is an oblique shock first of all we define the two first of all straight normal shock what is straight normal shock so far we have discussed that the shock with respect to a duct it may be flow past a body external flow also the shock will be there normal shock is a shock straight means this is straight there is no curvature in the shock and this is very thin the dimension the dimension is the order of the molecular dimensions thin straight and normal shock means the velocity of approach v 1 and the velocity behind the shock after the shock both are normal to the shock wave that means this angle included angle is 90 degree and since they are both of them are normal to the shock wave they have in the same directions that means both are in the same directions and which is normal to the shock wave which is straight normal shock wave now if I show the same thing we do not consider a curved shock wave but straight shock wave straight but not normal oblique shock the basic difference from the not oblique shock from the normal shock is that now I am not drawing any duct I am drawing any that the shock wave is not perpendicular to the incoming velocities first of all this is the first one v 1 that is incoming velocity is not perpendicular to the shock wave it makes an angle with the shock wave let this angle is beta and more over when it passes across the shock through the shock then what happens that this changes its direction v 2 change this is also not normal to the shock and this deviation in its path is given by delta this angle beta is known as shock wave angle that is the angle that incoming velocity makes with the thin straight shock which is oblique in nature because there is not 90 degree and the velocity which goes out that means the velocity after the shock is not in the same direction of its of that before the shock there is a change in the direction in case of an oblique shock wave and that is known as delta so delta is the angle of deviation that is the angle by which this velocity deviates their magnitude obviously this is acceleration that is ok angle of deviation this is known as oblique shock well now I will we will analyze the oblique shock so now with this concept in the background as the definition of the straight oblique shock now what I will do you see that let us consider a oblique shock like this and let us consider the oblique shock a control volume very small control volume as we did for normal shock for derivation of all equations you remember fundamentally by application of the conservation equations of mass momentum and energy the same thing now I will do so I take a control volume ok this is a control volume cv this is a control volume ok now this is my incoming velocity v1 and this v1 is having an angle this angle beta shock wave angle but this can be resolved into two component one is along the shock wave that is vs1 and another is perpendicular to the shock wave vn1 at the inlet n is the normal s is the along the distance that is usually presented as s now the outlet velocity here it has got a deflection so this is the deflection delta ok it has also can be it is also can be resolved into component vn2 and vs2 this angle is delta this is the angle of deviation and this is beta well now you see that what we do first consideration is that before applying any conservation equations we will tell you that since we know along the shock the fluid properties do not change it changes only across the shock so therefore there is no force acting along the shock there is no change in momentum which means that if we take a very infinite small control volume this vs1 and vs2 is same this is very very important assumptions of the oblique shock and this is the key assumption you will see after what the entire analysis depends upon that that means they have a velocity v1 they have a velocity v2 outlet outlet means after the shock but they are such a way that tangential component of the component along the shock wave are equal so only inequality is that vn1 and vn2 they are not equal so this is first thing we have to remember now if you write the therefore there is no change in momentum in this direction the force acting in this direction is 0 now if we write the continuity equation mass flux rho since vs1 is vs2 they cannot carry any extra mass additional mass in the control volume so control volume mass can be written if this is the cross sectional area and for the same cross sectional area the continuity equation will be rho1 vn1 this is 1 this is 2 so the properties are like that p1 rho1 t1 and p2 rho2 and this concept will remain same that p2 will be more than p1 rho2 will be more than rho1 because it is a compression process across the shock t2 will be more than t1 but now you just see how in a very simple way we did this is the conservation of mass or mass continuity now let us write the momentum theorem that means the equation of motion that is the momentum theorem that is the momentum theorem with respect to this control volume now what is the rate of net rate of momentum a flux from the control volume in the normal direction because in the axial direction it is 0 we have already discussed so this will be the mass flow rate is this this is the mass flow rate so mass flow rate into vn1 that means it will be net momentum a flux will be rho2 vn2 into vn2 so therefore vn2 square minus when here I will write the mass as rho1 vn because when I write the mass with the outlet velocity or the velocity after the shock I write this one and when at the inlet or at the approach I write this one this is the standard practice so this is the net rate of momentum a flux that is what momentum momentum normal direction momentum a flux from the control volume and under steady state this will be the net force acting on the control volume the normal direction and that is nothing but the pressure forces that means this becomes is equal to p1 minus p here the area terms cancels out because this is the mass flow rate per unit area I am sorry this is the mass flow rate per unit area so therefore area cancels from both the sides so this is your equations in the this is a momentum equation this is the momentum equation now if we write the energy equation what is the energy equation you write now the energy equation energy equation you write that h1 plus v1 square by 2 neglecting the kinetic and potential energy as you know the the similar treatment the shock across the shock the process is adiabatic so therefore we can write the energy equation without any work and heat interactions h1 plus v1 square by 2 is h2 plus v2 h1 is what cp t1 ok plus v1 square by 2 is cp t2 plus v2 square by 2 now the cp can be expressed as gamma by gamma minus 1 rt1 is sometimes this I did earlier also if you remember in the normal shock wave deductions this is was done this way the equations in compressible fluids are expressed in terms of only p and rho here now this v1 again can be written at vs1 square plus vn1 square and this must be equal to gamma by gamma minus 1 p2 by rho 2 plus vs2 square plus vn2 square now as our assumption will by 2 good very good by 2 by 2 so this and this cancels so therefore here also we can write the equation this 2 if you write like this 2 gamma by gamma minus 1 p1 by rho 1 minus 2 gamma by gamma minus 1 p2 by rho 2 is nothing but vn2 square minus vn sorry sorry vn1 square so this is the extract from this now you see since vs1 equals to v1 vs2 is the vs1 equals to vs2 then you see that automatically they have got no contribution if you now compare this with the normal shock equations you see instead of v1 and v2 v2 and v1 here also v2 and v1 we are just substituting the normal component of velocities that means all the relations of normal shock will hold good if we just change the actual velocity by its normal component so therefore this is clear and the other relationships that which we get in hook in your relationship is the which we get that p2 by p1 rho 2 by rho 1 rather rho 2 by I am sorry rho 2 by rho 1 let me write that again the rank in hook in your relations that we already deduce for this normal shock rank in hook on your relation rank in hook on your relation remains the same that means rank in hook on your relation is rho 2 by rho 1 in terms of p2 by p1 usually this is expressed in terms of p2 by p1 I write it for your recapitulation here only just recapitulation I write you plus 1 divided by gamma plus 1 by gamma minus 1 plus p2 by p1 so this is one relation so this relation is not altered because this is the relationship between the scalar quantities t2 by t1 is gamma plus 1 by gamma minus 1 into p2 by p1 divided by gamma plus 1 gamma minus 1 plus p1 by p2 so this is the thing so this remains as it is so this just for your recapitulation now I come again to this same thing that now obviously a intuition tells that when you compare this if you just substitute this normal velocity you feel that all the conservation equations for normal shock and oblique shock remains same since vs1 is equal to vs2 now if we want to do with the Mach number because what we want we want a relationship with the Mach number then what we will see that vn1 now before that we are just like to write one thing vn1 here itself that vn1 what is vn1 vn1 we can write vn1 we can write this one sign this is beta that means this is beta the vn1 is v1 sign beta what is vn2 vn2 is this one so this will be beta this will this is beta minus delta v2 sign this angle is beta minus delta why because this is beta and this is delta so therefore this angle is beta minus delta because this angle this angle is this angle my drawing is not in scale these are two parallel line so this angle is what this is beta with this this is the shock and this is the original line and this is the deviation delta so this is beta minus these two angle beta minus delta you can make it out by yourself v2 sign beta minus that means if I substitute instead v1 v1 sign beta and v2 v2 sign beta minus delta then this exactly matches with the relationships of the normal shock and then the most important relationships which we want that will come out that is what p2 by p1 you remember that was in terms of the approach mach number upstream mach number what is that it was m1 so now since this is with the normal component so automatically m1 will be m1 sign beta actual m1 is v1 by a1 so m2 is equal to m2 sign beta minus delta what is this m1 m1 sign beta does not mean that sign beta is equal to 1 that means in m upstream this m1 is m upstream actual m1 sign beta that means what we will do that is this is m1 sign beta rather you can forget this it will give confusion that if we use m1 sign beta as the approach m1 and m2 sign beta minus delta is the mach number as the mach number after the shock wave you can use the same relationship of the normal shock that means the normal shock relationship the m1 should be replaced by m1 sign beta and m2 should be replaced by m2 sign beta where m1 m2 is the actual mach number here for the oblique shock oblique shock mach number is v1 by a1 and a1 is root over gamma rt1 and this mach number is v2 by a2 there is no doubt this is the definition of the mach number and a2 is root over gamma rt2 so if we just normalize this mach number by sign beta and normalize this mach number by sign beta minus delta because of this logic that we have seen since vs1 is vs2 then we can use this m1 sign beta in place of m1 in normal shock relation and replace m2 sign beta minus delta as the m2 in normal shock relations that the same relationship which you will get for the oblique shock now it was 2 gamma m1 square so this was m1 square sign square beta minus gamma minus 1 divided by so in case of normal shock p2 by p1 was twice gamma m1 square minus gamma minus 1 divided by gamma plus 1 similarly if you write rho2 by rho1 that will be gamma plus 1 into m1 square but here it will be gamma plus 1 into m1 square sign square beta that means m1 for normal shock wave for normal shock wave is m1 sign beta for oblique shock wave oblique similarly m2 for normal shock wave will be replaced by m2 for normal shock wave m2 sign beta minus delta for oblique shock wave oblique shock wave for oblique oblique oblique shock wave that means we are replacing this and if we do that you will get the required relationship for the oblique shock to gamma minus 1 and this was m1 square that means this is again m1 square sign square beta similarly if you recollect last class I did t2 by t1 equals to what t2 by t1 equals to 2 plus gamma minus 1 into m1 square so this will be m1 square sign square beta and 2 gamma into 2 gamma m1 square that will be m1 square m1 square sign square beta minus gamma minus 1 and denominator was gamma plus 1 whole square gamma plus 1 whole square m1 square that means m1 square sign square beta now another interesting thing is that there was a relationship between m2 and m1 so if m1 is given I can get all parameters p2 by p1 that change in the property along with the m2 that is the mach number after the shock here e2 that relationship m2 has to be again replaced by m2 sign beta so actual relationship was m2 square equals to here m2 square sign square beta minus delta equals to what parallelly you see that if you remember that equation for the normal shock this will be m1 this was m1 square this is m sign square beta plus 2 by gamma minus 1 divided by 2 gamma m1 square so m1 square sign square beta divided by gamma minus 1 minus 1 this is the so if you see that expression compare with the normal shock only m1 is replaced by m1 sign beta and m2 is replaced by m2 sign beta so this is the thing that means now I can tell you that a oblique shock that a oblique shock again this is a very interesting thing that a oblique shock which has got two component which is striking the velocity v1 with the shock angle beta which is striking with the v1 value has got a normal component and the tangential component and this is being deflected with an angle delta v2 which also has a normal and the sorry my drawing is like that vs2 this is parallel to this and vn2 this is vs2 with the premise that vs1 is vs2 can be considered equivalent to that means if we now attach a axis just you think which is moving with vs1 and vs2 along the shock wave that means vs1 is vs2 is vs so if we attach an axis or if we make this representation respect to axis which is moving with the vs velocity that means this magnitude in this direction this is nothing but a shock wave normal shock wave that means this is like this that means this thing is with this premise is resolved to this this can be reduced to this with respect to a coordinate system which is moving with vs with respect to that coordinate system this is a relative velocity v1 with respect to vs which is nothing but the normal component and similarly with respect to the coordinate system which is moving with vs the velocity v2 relative to vs will be vn2 clear so therefore this will be resolved to a normal shock wave with vn1 and vn2 and all normal shock wave relationships will be valid provided the actual Mach number is normalized with sin beta and here sin beta minus delta so therefore these two things that beta and delta are very important parameter shock wave angle and the angle of deviation so therefore this is clearly reduced to a normal shock wave with the normal component of velocity and with a modified Mach numbers now with the similar treatment of the normal shock wave from the entropy considerations first we can tell m1 sin beta has to be greater than 1 because this has to be greater than 1 means this has to be supersonic that means the flow approaching the shock has to be a supersonic one so therefore this is one very important criteria which means that from here one can write equals to rather it can be at the limiting case equals to 1 sonic that means sin beta will be equal to greater than 1 by m1 now you see this gives a very important information that is the minimum value of beta minimum value or rather I write the sin beta sin or rather beta beta beta minimum is equal to sin inverse 1 by m1 what is this this is the Mach angle you remember which was told earlier this is the Mach angle that means the minimum shock wave angle is the Mach angle 1 by m1 what is the maximum shock wave angle that is the maximum value that a sin may have that is equal to pi by 2 that means the maximum value is the normal shock and the minimum value is the Mach angle that means the oblique shock limit is between Mach angle and the normal shock when 1 by m1 is the Mach angle if you just now substitute the Mach angle sin beta beta is sin inverse 1 by m1 sin beta is 1 by m1 you get p2 by p1 is equal to 1 if you put that m2 this is 1 2 gamma minus gamma plus 1 that is gamma plus 1 by gamma plus 1 that means in that case p2 by exactly the relationship which we got for a Mach wave that means the this is the Mach wave that means the oblique shock wave in this case becomes a Mach wave in one limiting case and in another limiting case with the maximum value of the shock wave angle it becomes totally the normal shock wave like this so this is one thing next is that again from the same thermodynamic consideration we showed earlier because it is now reduced to a normal shock sin beta minus delta has to be less than 1 that means we know that flow is subsonic now this is very interesting we saw that m2 is less than m1 and m1 has to be greater than 1 that means therefore sorry m2 be less than m1 and sorry m2 is definitely less than 1 m2 less than 1 and m1 has to be greater than 1 m2 has to be less than 1 sorry m2 has to be less than 1 automatically m2 is less than 1 so these two things we proved we proved these two things that a shock wave is such that supersonic flow is changed to subsonic flow so therefore Mach number is less than 1 but here you see in oblique shock wave m2 sin beta minus delta will be less than 1 for the same considerations if you go back to the normal shock relations and their analysis which means that m2 will be less than 1 by sin beta minus delta since the sin beta minus delta is less than 1 always this is not necessarily that always m2 will be less than 1 that means m2 sin beta minus delta less than 1 that means in that case m2 may be greater than 1 depending upon the value of sin beta minus delta so this is a very important consideration now m2 is actual Mach number sometimes it is asked if you are asked that if there is a oblique shock wave a supersonic flow after the oblique shock wave can remain supersonic yes it may remain because the limitation is m2 sin beta minus this depends upon the shock wave angle and the deviation angle of deviation this is the restriction that the reduced Mach number has to be less than 1 so the actual Mach number may or may not be less than 1 it may be less than 1 or it may not be less than 1 and some circumstances it may be greater than 1 depending upon the value of the sin beta minus delta so this is one very important considerations now we will develop one time is there now we will be develop one expression now we are interested to develop one expression between beta delta and m1 why you will understand afterwards let us make some algebraic calculations to do it now you see that what is beta now beta is this one so beta 1 beta 1 beta 1 so beta 1 is vn1 by vs1 tan beta 1 sorry similarly this is beta minus delta because this is delta and this is beta this is shown in another diagram beta minus delta so tan beta minus delta is vn2 by vs2 so let us write that that tan beta is vn1 by vn1 this is vn2 by vs1 and tan beta minus delta is equal to vn2 by vs2 now sorry if you make a ratio then tan beta minus delta divided by tan beta is equal to vn2 by vn1 and from continuity what is this vn2 by vn1 is rho1 by rho2 now this rho1 by rho2 is again can be written is this rho1 by rho2 can again be written in terms of your 2 plus gamma minus 1 as I have done earlier 2 plus gamma minus 1 m1 square sin square beta divided by that just now I have done gamma plus 1 into m1 square sin square beta at the present moment let I consider this as x then tan beta minus delta is tan beta minus tan delta divided by what 1 plus tan beta tan delta this is tan beta minus delta this is tan beta minus delta that divided by tan beta so therefore LHS this thing becomes what 1 minus tan beta tan alpha by tan beta tan alpha by tan delta sorry tan delta by tan beta divided by 1 plus tan delta tan beta equals to x now if you put the value of x and make little adjustment then first of all you put the value of if you clear this thing this thing will come as tan if you just make a rearrangement then you will get tan delta is 1 minus x x is what x is this one x is this one 1 minus x into tan beta divided by x tan square beta plus 1 and if you put the value of x finally you get an expression that with some rearrangement tan delta is that is important one 2 cot beta this will not come very fast next step but you will have to make 2 3 algebraic simplification but this will come very straight forward no other relationship will be required only you have to know the trigonometric identities that means tan delta this is coming clearly from here one line next step it comes here now when you substitute this x we replace this as x here this will be little complicated one in a sense it will be big one and if you make a rearrangement and clear it ultimately you get tan alpha is 2 cot beta m 1 square sin square beta minus 1 2 plus m n square into gamma plus cos 2 beta this is one very important relation which expresses the angle of deviation in terms of the shock wave angle and the approach mach number upstream mach number m 1. So, I will continue this in the next class since the time is up out and I think you do not have time thank you for today.