 We will continue with our discussions on compressible flows. We will now discuss on something known as stagnation properties. This term stagnation we have come across during our earlier topics that we have covered in fluid mechanics and one of the literal meanings is that the fluid is brought to rest at a point so that the velocity is 0. So, stagnation point is a point where the velocity of flow is 0, but that does not mean that it suffices the requirement of description of stagnation properties. So, we will now look into more deeply about some of the important properties which dictate the nature of the process by which the stagnation state is achieved. To do that we will refer to first an important thermodynamic property and when we are referring to a thermodynamic property we will be basically referring to the first law of thermodynamics to specify that. So, if we say that you have a control volume say some control volume which has some inlet I and some exit E. Let us say that there is some rate of heat transfer to the control volume say q dot cv and say some rate of work done by the control volume as w dot cv and let us say that the flow is steady and the state or the properties of fluid within the control volume is steady that means the properties within the control volume do not change with time. So, if these 2 conditions are simultaneously achieved then the corresponding form of the first law of thermodynamics is like this where h is the property enthalpy which is the internal energy plus pressure by the density. Now this particular form is also known as a steady flow energy equation just for common understanding. So, this is nothing but the first law of thermodynamics expressed for a flow process across a control volume when the flow is steady and the states within the control volume are also steady. Not only that there are certain more important assumptions what are the important assumptions all the properties at the inlet and the exit are uniform that means the velocity profiles are uniform the thermodynamic properties like enthalpy those are uniform. So, it is like approximately a one dimensional representation where uniform properties across the cross section for the inlet and the outlet. Now we are interested to apply this equation for compressible flows. So, when we apply this equation for compressible flows first of all if you have a system where you are not having any mechanism of extracting work from that. So, then this rate of work done will be 0 by the control volume. Then if what kind of process we are considering we are considering an adiabatic process. So, if it is an adiabatic process that means heat transfer across the control volume is 0. We are not committing whether it is reversible or irreversible. So, the reversibility or irreversibility features when we talk about the second law of thermodynamics when we discuss about that. But in the first law of thermodynamics it is just good enough to say whether it is adiabatic or not irreversible or irreversible does not feature here. Now, in most of the high speed gas flows the effects of the thermal effects and the kinetic energy effects are more important than the changes in potential energy. So, that is negligible and therefore you are left with h i plus u i square by 2 is equal to h e plus u e square by 2, okay. Now, let us say that we are thinking about 2 sections instead of i and e some generic section where the where the enthalpy this is specific enthalpy that is enthalpy per unit mass is h. The velocity is u this is at some section and when you go to some another section that section is a special section when the velocity is brought to 0. So, this is a section of stagnation. So, we are interested to see that what is the corresponding enthalpy there and let us say that the name of enthalpy at that stagnation section is h 0. Since it is a one dimensional treatment section and point are the same it is basically uniform. So, we can say from here that h plus u square by 2 is equal to h 0 because at the stagnation u equal to 0. So, this h 0 is known as stagnation enthalpy for an ideal gas it is known that dh equal to cp dt this is for an ideal gas where cp is a function of temperature in general. But when we talk about a perfect gas cp is a constant. So, if we say perfect gas this is a constant for perfect gas therefore, we may say h minus h 0 for a perfect gas is cp into t minus t 0 where t 0 is the temperature corresponding to this stagnation enthalpy this is known as stagnation temperature. Why we refer to the temperature? Because temperature is a directly measurable quantity from experiments. So, it is important that we refer to that. So, this is known as stagnation temperature. So, we can write cp into t minus t 0 plus u square by 2 equal to 0. Now, we can write cp in terms of r and gamma because just recall that cp minus cv is equal to r and cp by cv is equal to gamma. So, you can write cp into 1 minus 1 by gamma equal to r that means cp is equal to gamma r by gamma minus 1. So, we can write t 0 minus t in place of cp we write gamma r by gamma minus 1 is equal to u square by 2. u square by 2 is what? You can write u by c is equal to the Mach number say m. So, u square is equal to m square into c square. Now, what will be the expression for c will depend on the nature of the process. If it is an isentropic flow c square is gamma rt. So, this is equal to m square gamma rt for isentropic flow. See till now whenever we define this stagnation temperature. So, the stagnation temperature how it is defined? The stagnation temperature is defined from this equation as t 0 is equal to t plus u square by 2 cp right. This definition does not require the process to be reversible. So, if you want to find out that what is the stagnation temperature. In fact, this definition does not require any process because it is just an expression. So, if you know the temperature and velocity at a point and specific heat at a point you know the stagnation temperature at a point. So, stagnation temperature is defined at a point irrespective of what is I mean what kind of state is there at that point. If you want to physically achieve a stagnation temperature you have to bring the process to rest at a point in an adiabatic manner. It need not be reversible that is not necessary because the definition followed from the first law of thermodynamics by setting heat transfer equal to 0 without imposing any condition of reversible process. So, keep one thing in mind stagnation temperature just like any stagnation property is a property. So, it does not mean that at one point if it is not a stagnation point it will not have a stagnation temperature. It will have a stagnation temperature because it is just dependent on the local temperature velocity and cp. So, it is just like a combination of properties and therefore, it is a property. If you say that I want to physically achieve the property then you have to follow this kind of a process adiabatic process. Now, if you want to use this expression u square equal to m square into gamma RT that means you are now imposing the additional constraint that it is an isentropic process that means a reversible process. So, for an isentropic process this will be m square gamma RT by 2. So, from this what follows is so, gamma into R will cancel from both sides. So, you will get T0 by T is equal to 1 plus gamma minus 1 by 2 m square. So, this is the relationship between the stagnation temperature and the temperature at a point. What are the assumptions under which this is valid? It is an isentropic process that you have to keep in mind otherwise the more general expression is this one. Now, you may relate the stagnation just like stagnation temperature you may have a stagnation pressure and stagnation density. So, those properties also you may find out. So, for an isentropic flow for isentropic process you have PV to the power gamma equal to constant that means basically in terms of T and P you can write T2 by T1 is equal to P2 by P1 to the power gamma minus 1 by gamma. So, just like you have related the stagnation temperature similarly you can relate stagnation pressure P0 by P is equal to T0 by T to the power gamma by gamma minus 1 that means 1 plus gamma minus 1 by 2 m square to the power gamma by gamma minus 1. This P0 is stagnation pressure remember assumption again is isentropic flow. Now, so this considers compressibility. We have earlier seen a case where we used Bernoulli's equation to define a stagnation pressure and that was without consideration of any compressibility. So, we use just the Bernoulli's equation Bernoulli's equation means it is an incompressible flow assumption and the stagnation pressure. So, if you consider the incompressible limit P0 is equal to P plus half rho u square neglecting the potential energy effect. What are the assumptions? It is an incompressible flow. We use the Bernoulli's equation that means we implicitly assume that it is a frictionless flow. So, if you want to achieve physically this P0 you have to bring the fluid to rest in a reversible and adiabatic process that will mean that it is a frictionless process. If you want to achieve T0 physically you have to bring the fluid to rest in an adiabatic process it need not be reversible whereas to achieve P0 physically you have to bring it to rest by ensuring both that it is reversible as well as adiabatic. This is a very important distinction between like how you achieve physically a stagnation pressure and a stagnation temperature. So, now when you come to this incompressible flow so here you can write P0 by P for an ideal gas P by rho is RT. So, this is 1 plus half u square by RT. Remember that if it is an isentropic flow C square is gamma RT. So, this you can write 1 plus gamma by 2 u square by C square u square by C square is M square. So, this is an expression which is valid for incompressible flow. This is an expression that is valid for compressible flow. It may be interesting to see that in a certain approximate limit these 2 equations in certain approximation they agree with each other. To do that what we may do this is the more general expression. So, we may expand this in a binomial theorem just like 1 plus x to the power of n. So, if we expand that in a binomial theorem so P0 by P compressible. So, 1 plus x to the power n is like 1 plus nx plus n into n minus 1 by factorial 2 into x square plus let us just write another term n into n minus 1 into n minus 2 by factorial 3 into x cube. So, this is 1 plus gamma by 2 M square. So, you can see that up to the first term it is just like the incompressible flow expression. The remaining terms are the corrections because of the compressibility effect. So, what are the corrections? So, you can write may be the first 2 corrections. So, 1 is gamma minus. So, this is gamma by 8 M4 right then plus this is 2 plus gamma into gamma by 48 M to the power 6 like that sorry 2 minus gamma. So, what is this important implication say you are using a p-tort tube to measure the flow velocity at a point and you are negligent of the compressibility effect and you are using this expression for P0 by P because in a p-tort tube you may get the difference between stagnation and static pressure by connecting a manometer between the stagnation point and another point which is a point located upstream to that. So, when you get that expression that expression will require a correction and the leading order terms that will dictate the correction the first 2 leading order terms are like this. So, if it has to have a compressibility effect these extra terms for correction need to be invoked and we can see this these terms are higher powers of M. So, as M becomes smaller and smaller this becomes more and more irrelevant that means for low values of Mach number the compressibility effects are smaller and smaller that is quite intuitive ok. Next what we will consider next we will consider the similar thing that we have discussed till now may be isentropic flow to begin with but through a variable area duct till now we have not taken area as a variable parameter but now let us generalize the previous discussion somewhat and consider isentropic flow through a variable area duct. So, when we say a variable area duct maybe it is something like this maybe it is something like this we will not specify what it is but the area of cross section will vary. So, what are the basic equations that we may use here one is of course the continuity equation and we are all in all cases we are assuming that it is a one dimensional flow. So, rho into A into U that is a constant this is from the continuity equation then the fluid flow equation here remember we are talking about an isentropic flow. So, reversible and adiabatic. So, it is having no friction that means it is an inviscid flow. So, what will be the governing equation for that Euler equation one dimensional form of the Euler equation. So, Euler equation so that is what dp by rho plus U du equal to 0 that is the differential form of the Euler equation. Remember that in all cases of compressible flow that we are discussing we are neglecting the changes in potential energy the gravity effects are negligible as compared to the other effects. So, this is like the these are sort of like the fluid flow equation and the thermodynamic constant is the energy equation. So, energy equation is like h plus U square by 2 equal to constant but first of all we will not consider this energy equation as an important consideration but we will concentrate on these 2 equations and see what we get out of that. So, continuity equation rho into A into U equal to constant. So, what we will do is we will take log of both sides. So, we will have ln rho plus ln A plus ln U is equal to ln constant and then differentiate means d rho by rho plus dA by A plus du by U is equal to 0. Say this is equation number 1 and say this is equation number 2. Now, if we want to relate these with the sonic speed see in equation number 1 there is d rho in equation number 2 there is dp. So, somehow if we find out dp d rho that will give c square. So, we can relate these behaviour with the sonic speed. So, we can so we write from 2 that dp is minus rho U du and or if you want to relate it with U square. So, dp by U square is minus rho U du by U square. So, minus rho du by U why we have written in this form is because you have another form du by U in equation 1. So, the let us say this is 3 and from equation 1 you have d rho is equal to minus rho into dA by A plus du by U say this is equation 4. So, if you divide equation 2 by equation sorry equation 3 by equation 4 you get dp d rho into 1 by U square is equal to du by U divided by dA by A plus du by U. If you consider it as an isentropic flow d p d rho is c square. So, for an isentropic flow the left hand side is 1 by n square because m is U by c. So, from here you can write dA by A plus du by U is equal to m square du by U which means dA by A is equal to 1 minus or is equal to m square minus 1 into du by U. This is the relationship that governs the change in area with change in velocity and it is one of the very important physical relationships which we have to carefully study. So, if you write say dA by A. So, when you write dA by A we think about 2 possibilities. See I mean one of the possibilities is that along the direction of the flow the area is increasing or area is decreasing. So, this is an example with dA by A greater than 0. This is an example with dA by A less than 0 assuming that this to be the direction of flow. Now, if you consider m equal to 1 as a case. So, m equal to 1 if you consider then that will imply what dA by A equal to 0. That means if you want to achieve m equal to 1 m equal to 1 is called as a sonic condition. In an isentropic flow it must conform to dA by A equal to 0. dA by A equal to 0 if you look into it mathematically it may be either maximum area or minimum area right. Physically we will see that it will it will conform to the minimum area. So, if you consider a possibility of these 2 types say you consider this as a case. Let us say that you have the inlet flow with a particular Mach number whatever the Mach number at the inlet may be greater than 1 or may be less than 1. Now, we are interested to see the consequence. If you consider this particular geometry if you see that no matter wherever m equal to 1 is achieved we will see later on with a physical example that how m equal to 1 may be achieved. See just think this geometry. So, this particular geometry even if it is an incompressible flow you see what is your target. The target could be to increase the velocity that is why you reduce the area of cross section. So, these types of things where you have a variable cross sectional area with an intention of increasing the velocity and therefore the consequences reduction of pressure that is known as a nozzle. So, if you have a nozzle your target is to have an increase in velocity, but just this geometry does not ensure that it will be like that for a compressible flow. For an incompressible flow area reduction will imply that the velocity will increase. For a compressible flow if you think it like that it may be wrong because rho is also another factor. It is rho into a into v not just a into v. So, rho may vary in such a way that it might be that if area is increasing velocity is also increasing. So, it all depends on the Mach number at which the fluid is there. So, we will see later on that what is the physical implication of this minimum area when we discuss more about the nozzles the converging nozzle. This is called as the converging nozzle and the converging diverging nozzle, but look it in the other way. If m is not equal to 1 then can you have dA by A equal to 0? Definitely, if m is not equal to 1 then you may have dA by A equal to 0 if dU by U equal to 0 that means U is either a minimum or a maximum. So, from this we come to a very important conclusion that if for the time being we take that at the minimum area m equal to 1 is achieved that does not mean that at the minimum area m equal to 1 is always achieved. It means that if m equal to 1 is achieved somewhere it has to be a minimum area location, but the minimum area location will not always have m equal to 1. If it does not have m equal to 1 it will have either a minima in m or a maxima in m and we will see that how it is possible. To understand that what we will do is we will consider two different cases. One is m greater than 1 another is m less than 1. So, if you consider m greater than 1 you can see that if you have a reduction in A that means if dA by A is reducing then dU by U that is also reducing. That means if area decreases the velocity also decreases if the flow is supersonic. This is non-intuitive incompressible flow consideration will say if area increases the velocity will decrease. So, why it is happens? Roughly if you see that what you are keeping fixed rho into A into U. So, you are reducing A you are expecting U to increase, but with a reduction in A maybe there is such a high increase in density that there is actually a significant reduction in U to keep the product as constant. So, that is what is happening here physically. So, what we get out of this is when m greater than 1 that means if you have dA by A that is if area is reducing then the velocity is reducing therefore it is not acting like a nozzle. It looks like a nozzle, but it is acting like a diffuser. Diffuser is something where you want a decrease in velocity and an increase in pressure. So, here it is acting like a diffuser. So, physically looking like a nozzle does not mean that it is a nozzle. One has to see what is the Mach number range in which it is operating. On the other hand if m is less than 1 you have dA by A will be if it decreases that will mean that dU by U is positive that means U increases. So, if so this situation we may intuitively think as physically analogous to an incompressible flow behaviour in a nozzle only if Mach number is less than 1 only if it is a subsonic flow. If it is a supersonic flow that is not the case. So, if it is a subsonic flow what happens if it is a subsonic flow so now think about such an arrangement. So, this is an arrangement where the area is reducing it comes to a minimum and then increases. So, in this kind of an arrangement if you see if you say if you plot so let us say that the arrangement is like this we plot the Mach number versus x for 2 cases assuming isentropic flow for 2 cases one is the entry Mach number greater than 1 another is entry Mach number less than 1. So, if the inlet Mach number is less than 1 say the inlet Mach number is less than 1. So, this is Mach number equal to 1 say inlet Mach number is half like this. So, as you go through the converging section the velocity will increase and therefore the Mach number will increase. So, the Mach number will increase and it will come to a maximum when dA by A is 0. So, it may not achieve 1 but it will come to a maximum and then it will fall and then it will fall because the area variation is different. On the other hand if the inlet Mach number is greater than 1 then the velocity will reduce in the converging section and therefore the Mach number will reduce it will come to a minimum therefore here and then will increase again. So, if the entry is supersonic the throat may have a minimum Mach number if the entry is subsonic the throat may have a maximum Mach number but these are limited by Mach number equal to 1. So, if it is supersonic it is remaining supersonic for all the locations it is coming only to a minimum here but that is still greater than 1. On the other hand if it is subsonic it is remaining subsonic everywhere at the throat it is coming to a maximum but still greater less than Mach number 1. So, from this we conclude a very important thing it is not necessary that at the throat this location of minimum area is called as the throat that at the location of the minimum area it is not necessary that we will always have Mach number equal to 1. But the converse is true if we have at some location the Mach number equal to 1 that must be at the throat. Again remember there are very important assumptions associated with it the assumption is it is an isentropic flow. So, if it is a flow with friction and flow with heat transfer then the location of this sonic point Mach number equal to 1 shifts it is no more at the throat it is at either the upstream or the downstream depending on the heat the heat transfer in the flow and friction in the flow. Now, it is also possible to write this expression of dA by A exclusively in terms of dM by M. So, let us try to do that because it will give an exclusive variation of or exclusive relationship between Mach number and the area. So, let us write say d so, we know that the Mach number is equal to the C into U. We are interested to express dU by U in some way in terms of dM by M. So, that we write dA by A explicitly in terms of dM by M sorry U equal to C into M right. Now, what you may do is you may take log of both sides and then differentiate see for all these cases by this time you have realized that there is a purpose in taking the log and differentiating because you are getting this type of expression dA by A dU by U like this by taking log and differentiating. There are other ways of getting it, but this is the easiest way of getting such forms. So, log of U equal to log of C plus log of M and then you differentiate. So, you will get dU by U is equal to dC by C plus dM by M. So, in an effort of expressing dU by U in terms of dM by M we have come up with a new unknown dC by C, but C may be expressed in terms of the temperature. So, let us give this equation some numbers. Let us say this is equation number 5 and say this is equation number 6. Now, we know that for an isentropic flow see we are writing all these equations for isentropic flow. So, C is equal to root over gamma RT for isentropic flow. So, again we take log. So, log C is equal to half log gamma plus half log R plus half log T that means dC by C is equal to half of dT by T. So, dC by C is half of dT by T therefore, you can write from 6 that dU by U is equal to half dT by T plus dM by M. Again another new unknown dT by T has appeared, but till now we have not used the energy equation that we may use. So, energy equation what it was? H plus U square by 2 equal to constant that was the energy equation. So, you have dH plus U dU equal to 0 and for a perfect gas this is Cp dT plus U dU equal to 0. In place of Cp we can write gamma R by gamma minus 1. If you want to write it in terms of dU by U just divide both the terms by U square and U square is equal to M square into gamma RT. So, this is M square gamma RT. So, gamma R gets cancelled. So, 1 by gamma minus 1 M square dT by T plus dU by U equal to 0. So, we have been successful in writing dT by T in terms of dU by U. So, from this we get dT by T is equal to minus of gamma minus 1 M square into dU by U. So, this is 9. So, if you combine 8 and 9 this is dU by U dT by T is written in terms of this half of the right hand side and so that will become 1 plus gamma minus 1 by 2 M square dU by U is equal to dM by M. So, that we may substitute in equation number 5 which is there to get dA by A is equal to M square minus 1 into M square minus 1 into M square minus 1 into 1 plus in place of dU by U we may write dM by M. So, from this equation we may get an explicit relationship between the area and the Mach number. So, if you know how area varies with x you can easily get how the Mach number varies with x. So, for a given geometry A as a function of x you will get an output as a Mach number as a function of x and from here you can clearly verify the statement that we have just made. So, if M is greater than 1 see the denominator is always positive because gamma is Cp by Cv Cp is greater than Cv always. So, this is positive M square is positive. So, denominator is positive. So, numerator is dictating the sign. So, if M is greater than 1 dA by A positive means dM by M positive and if M is less than 1 dA by A negative means dM by M positive. So, whatever statements we made in terms of dU by U similar logic holds for dM by M and hence these plots you can verify the natures of these plots. Now, the other consideration regarding these types of flows is the achievement of a sonic condition. So, we have given some emphasis on that and we have seen that if the sonic condition at all has to be achieved that is achieved at a location where dA by A is equal to 0. Although at dA by A equal to 0 sonic condition need not necessarily be achieved, but if it is achieved at all it is achieved there. Now, when the sonic condition is achieved these sonic states this is usually given by a quantity star in the description of the flow. So, this is just symbolic way of representing. So, if at a state sonic condition is achieved let us say the corresponding velocity is U. We call it U star the corresponding pressure is P we call it as P star corresponding density as rho star corresponding area as A star like that. These star quantities are important because these are some important reference quantities. Remember that there may be cases when no sonic condition is achieved at all look at these examples either the top curve or the bottom curve sonic condition is achieved nowhere in the flow. So, still U star rho star P star rho star A star these quantities exist because these are hypothetical reference conditions. For example, A star is what? A star is at a given condition what could be an area at which a sonic condition would have been achieved not that that area is physically there in the flow. So, for corresponding to each state there is a corresponding start quantity and if you see tables of properties of compressible flows these start properties are there. So, these are very important reference quantities not that these values have to exist in the particular flow condition but these are reference quantities with respect to which other quantities may be expressed. For example, if you have say a condition where this sonic state is achieved then you may write rho A v or rho A u is equal to rho star A star U star. So, this is a hypothetical condition if the sonic state is achieved it may be achieved in reality may not be achieved in reality but if it were achieved then still it should satisfy the mass conservation. So, you may write as A by A star as rho star by rho into U star by U. Remember that U is equal to m into square root of gamma RT right for an isentropic flow. When it is U star it is square root of gamma RT star because m star equal to 1 star condition is sonic condition. So, when you write star that means the Mach number at that state is 1. So, this may be written as rho star by rho into U star by U is root over T star by T into 1 by m right. How you can write T star by T? We may use the stagnation properties and we may use the reference with respect to the stagnation properties. Remember we had T 0 by T is equal to 1 plus gamma minus 1 by 2 m square we just now derive this. If it is a stagnation property remember that when it is an isentropic flow the stagnation property does not change. Why? For example, the stagnation temperature does not change for an isentropic flow. Why it does not change for an isentropic flow? See consider the energy equation h plus U square by 2 this is a constant and this is equal to h 0. So, whatever may be if h varies U also varies in such a way that sum of these is equal to a constant and therefore, h 0 is a constant h is like C p into T for a perfect gas and that is why T 0 remains a constant. In fact, it is not necessary that it has to be isentropic just for adiabatic flow T 0 is a constant because reversible condition is never used here. But when you use this relationship it uses the isentropic condition also because it uses p by rho to the power gamma equal to constant for the sonic speed derivation from which this expression comes. So, T 0 equal to constant that means you can write T 0 by T star as what? What is T 0 by T star? 1 plus gamma minus 1 by 2 that is gamma plus 1 by 2. So, you can write T star by T by dividing these two expressions. So, T star by T is 1 plus gamma minus 1 by 2 m square by gamma plus 1 by 2. Also, we know that T by rho to the power gamma minus 1 is equal to constant for an isentropic flow. So, you can write T by rho to the power gamma minus 1 is equal to T star by rho star to the power gamma minus 1. So, rho star by rho is equal to T star by T to the power 1 by gamma minus 1. So, a by a star. So, you have T star by T to the power 1 by gamma minus 1 into T star by T to the power half. So, T star by T to the power half plus gamma minus 1. So, half plus sorry 1 by gamma minus 1. So, gamma plus 1 by gamma minus 1. So, this is T star by T to the power of gamma plus 1 by gamma minus 1 1 by m. That means, you have a by a star sorry one half is there to the power gamma plus 1 by 2 into gamma minus 1 this into 1 by m right. So, what it tells is that given a value of m you can get one a by a star. Looking from the other angle given one a by a star you could have some value of m, but it is not a unique value of m. You could have multiple values of m that is very clear from this because it is an equation having multiple roots. So, it will have basically two acceptable values of m and if you look into the tables of properties of compressible flows usually these quantities are referenced. So, you have for each Mach number certain important properties are given. So, if you look into isentropic flow tables which in the next class we will look more carefully. So, if you have the isentropic flow important properties what are the important properties? So, for each Mach number you can have p by p0, t by t0, rho by rho 0 and a by a star. It is important to tabulate this because if you know the Mach number you can calculate all others, but if it is a reverse problem that is if you know a by a star calculate what is the Mach number all those multiple roots you may have to solve non-linear equations, but if you have a tabulated set of data you can just read from the table and that is why the compressible flow tables are there with every chapter every book in the chapters of compressible flows the corresponding there will be an appendix where these properties are there. So, these properties will relate everything with a reference star the reference is either the stagnation state or the sonic. Important thing is that the stagnation state is the same so long as it is isentropic flow. If the isentropic nature of the flow is disturbed then the stagnation property changes that means t0 is same so long as it is isentropic, but if the isentropic nature is changed it is if it is no more isentropic then the stagnation properties change and one of the important mechanisms that can create such abrupt change in stagnation property is a shock is the presence of a shock wave. So, in our next class we will see that how these properties get changed when you have an abrupt discontinuity in a compressible flow medium present in the form of a shock wave. So, that we will take up in the next class. Thank you.