 In our previous lecture, we were discussing about the converging-diverging nozzles and let us work out an illustrative example on that. The problem is as follows. A convergent-divergent nozzle is designed to expand air from a chamber in which P0 equal to 800 kPa which is the stagnation pressure T0 equal to 40 degree centigrade to give the Mach number equal to 2.7 at the exit. The area of the throat is 0.08 meter square. This is the given data. So, the following questions are to be answered. Number 1, what is the area of the exit? Number 2, what is the designed mass flow rate? Number 3, what is the lowest back pressure for which there is only subsonic flow throughout the nozzle? Only subsonic flow throughout the nozzle. Number 4, what is the design back pressure? Number 5, what is the back pressure for which there is normal shock at the exit plane? Number 6, what is the back pressure for which there is no shock inside the nozzle? Number 7, what is the back pressure range for which there is oblique shock at the exit plane? Oblique shock at exit. And number 8, what is the back pressure range for which there is expansion wave at the exit? So, this is the set of questions that we would like to answer. So, let us look into this one by one. First of all, what is the area of the exit? So, to know that if we refer to A by A star, we know that A by A star is a function of Mach number only. This we have derived earlier. So, based on that, we can find out what is AE because the Mach number at the exit is given which is 2.7. A star is nothing but area of the throat because the sonic condition here is reached at the throat. So, from here we can get what is the value of AE which is 0.255 meter square for this problem. Second part, what is the design mass flow rate? Remember the design mass flow rate corresponds to the star condition that is m dot design is nothing but rho star A star into u star. So, individually we have to calculate rho star A star u star. So, how do we calculate rho star? Maybe we may calculate t star or p star. So, to calculate t star, let us use any of the isentropic relationships that is t by t star or say t0 by t is equal to 1 plus gamma minus 1 by 2 into m square that we know. So, we can find out what is t star by noting that when the star condition is there, it is m star and m star is equal to 1. So, putting that value, we may find out what is t star. Similarly, we may find out p star, rho star etc. by using the isentropic equation of change. Now, how do you calculate u star? Remember that u is equal to m into c which is square root of gamma rt. So, u star is equal to m star square root of gamma rt star with m star equal to 1. So, if we know t star, we can get u star and A star is nothing but equal to A throat. So, from all these considerations, by substituting the values, you can calculate what is the m dot design which for this case will come out to be 146 kg per second, third part. What is the lowest back pressure for which there is only subsonic flow throughout the nozzle? So, if you look at this pressure diagram for or pressure profile for different back pressures, we can see that the back pressure corresponding to the point c shown in the diagram is the back pressure, lowest back pressure for which there is only subsonic flow throughout the nozzle. So, how do you calculate that one? You know p by p0 is a function of the Mach number. And you know that you may also write it as A by A star as some another function of Mach number, say f1 Mach number. So, when you know that what is the area of the exit Ae by A star, Ae we have already evaluated, it is a function of Mach number at the exit. This has two isentropic solutions. One is corresponding to the point c, another is corresponding to the point g shown in the figure. Out of these two, one is the subsonic solution which corresponds to c, another supersonic solution corresponding to g. So, we get the subsonic solution from this, say you can get Me subsonic. Remember that 2.7 is the supersonic solution. So, we can get a subsonic solution also from the A by A star by referring to the isentropic tables. So, once you get that one, you may plug it back here to get p as a function of p0. So, that p will be the lowest back pressure for which the subsonic flow exists throughout the nozzle. So, we can find out this one to be equal to that is the pressure. At this point c as 780.5 kilopascal. Part 4, what is the design back pressure? The design back pressure you may calculate exactly in the same way as part 3, but you replace the subsonic solution with the supersonic solution. So, here m corresponds to the supersonic solution that is 2.7 that is pressure, the back pressure that is pressure at the point g in the figure by p0. So, from that you can get what is the pressure at the point g which is the design pressure and which may be evaluated as 34.4 kilopascal. Next, let us refer to the part 5. What is the back pressure for which there is normal shock at the exit plane? So, normal shock at the exit plane means we are referring to a condition which is like the back pressure corresponding to e. So, just upstream of the shock it was expanding like an isentropic flow. So, the Mach number was 2.7. What is the Mach number at the downstream of the shock? You get you know what the relationship between m2 by m1 for the shock. So, from the value of m1 you can get what is the value of m2 which is just at the downstream of the shock. So, once you know that what is the value of m2? You also know what is the value of p2 by p1 shock as a function of m2 and m1. So, if you know the value of m1 and m2 which you know from the previous step you can find out the ratio of the pressure downstream and upstream of the shock. And from there you can find out the pressure at e because you know what is the pressure at g. So, p2 by p1 is as good as pe by pg in the figure. So, from here you can find out what is pe. So, sorry what is pe? So, that value of pe for this problem the answer will be 2.7. So, we know the pressures at points C, E and G these we have already evaluated and the remaining 3 parts the answers to the remaining 3 parts of the problem may be given on the basis of this. Part number 6 what is the back pressure for which there is no shock inside the nozzle. So, we can see that the shock occurs outside the nozzle when the back pressure falls below the pressure of the point E. So, when pe is less than pe then there is no shock inside the nozzle. Part 7 what is the back pressure range for oblique shock at the exit? So, you can see that if the back pressure is between the points E and G marked in the figure there will be oblique shock inside the nozzle. So, the back pressure has to be between pe and pe and part 8 the back pressure range for expansion wave at the exit when the back pressure falls below the design pressure that is when pe back is less than pe G for example the point H as shown in the figure you will get expansion wave at the exit. So, depending on the back pressure which is there you may have different interesting phenomena at the exit and within the nozzle itself. Now till this time whatever we have discussed on the compressible flows through darks or nozzles we have assumed that the flow is isentropic. In other words we have assumed that the flow is reversible and adiabatic. However it is not always true that the flow will be reversible and adiabatic. In reality hardly there is a case when the flow is reversible and adiabatic. Therefore we need to consider situations when more general types of flow occurred. So, the generalization may be there with certain possibilities one is instead of the flow being adiabatic it may be a flow with heat transfer at the wall or and because of friction the flow becomes irreversible and that means that you could consider a case when the flow is irreversible and non adiabatic. In this particular course we do not have the possibility or scope of discussing all the detailed aspects of irreversible and non adiabatic flows together. So, we will consider a specific example where we have a deviation from the isentropic flow in the sense that the flow is adiabatic but irreversible because of friction in the flow. So, we will consider the adiabatic flow with friction and for simplicity we will consider the flow in a constant area duct. So, schematically the situation is like this you have a duct of say area A and some compressible flow is occurring through this duct the flow direction is along x and let us write the basic governing equations by assuming a one dimensional flow for such a case. So, the first basic equation is the continuity equation. So, what we get from the continuity equation rho into A into U that is equal to constant. So, rho into A into U equal to constant which means remember here rho is also a constant sorry here A is also a constant because it is a constant area duct. So, it boils down to rho into U equal to constant. So, we may take log of the expression and then differentiate to get d rho by rho plus dU by U equal to 0. Let us say this is equation number one the next fundamental equation that we may think of is momentum conservation. So, let us try to write the momentum conservation equation to do that let us consider a control volume of width dx the pressure acting on one side of the control volume is P. So, the force is P into A pressure acting on the other side of the control volume is P plus dP. So, the force is P plus dP into A the velocity is on the 2 sides of the control volume. So, on one side the velocity is U let us say on the other side the velocity is U plus dU. The other forces that act on the control volume are the frictional forces because of the wall shear stress. So, if tau wall is the wall shear stress it is the tau wall times the wall shear stress. Area on which it acts as the total shear force what is the area on which it acts let us say it is a circular duct as an example. So, if we consider this as a circular duct the area on which it acts is 2 pi r into dx. What is 2 pi r? 2 pi r is the perimeter of cross section. So, in general we can say that it is tau wall into P into dx where P is the perimeter of cross section of the duct. So, we can write a momentum conservation principle by using the momentum theorem as a resultant force as P into A minus P plus dP into A minus tau wall into P into dx. This is nothing but equal to m dot into U plus dP into dx. So, m dot into U plus dU minus U and m dot is rho into A into U. So, rho A U dU. So, from this it follows that if you divide all the terms by A. So, you have dP plus tau wall into P by A dx plus rho U dU is equal to 0. To write in a more compact form we may divide all the terms by rho U square. So, it becomes dP by rho U square plus tau wall by rho U square into P by A dx plus dU by U equal to 0. Let us say this is equation number 2. The next equation that we may write let us consider the energy conservation. So, for that we are basically writing the first law of thermodynamics for a steady flow process. There is no heat transfer because it is an adiabatic flow given. So, we have h plus U square by 2 is equal to constant. So, we can write dH plus U dU is equal to 0. Remember that for an ideal gas dH equal to Cp dT plus U dU equal to 0. Again we may write it in a compact form by dividing by U square. So, you have Cp dT. Let us write this as equation number 3. We will have a further simplification of equation number 3 or maybe we will call the next step as equation number 3. So, Cp dT by U square plus dU by U equal to 0 by dividing by U square. Remember that U square is nothing but m square into gamma RT and accordingly we may write in the next step that using Cp by R is gamma by gamma minus 1. So, this becomes dT by m square into gamma minus 1 in T plus dU by U equal to 0. Let us say this is equation number 3A. So, this is the energy equation. What other basic equations we have? We have equation of state P is equal to rho RT. So, we may again take log of both sides and differentiate to get dP by P minus d rho by rho minus dT by T equal to 0. Let us say this is equation number 4. And the fifth equation that we may get is the relationship between the sonic speed and the Mach number. So, U square is equal to m square gamma RT. This is a property relationship. So, from here we may differentiate, we may again take log of both sides and differentiate to get 2 dU by U is equal to 2 dm by m minus dT by T equal to 0. So, this is equation number 5. Sorry, equal to 2 dm by m plus dT by T. This is equation number 5. Now, we have from equation 1 through equation 5, 5 independent equations and these equations have their own unknowns. What are the unknowns? You have dP, d rho, dT, dm and dU. So, by eliminating variables, it is possible to obtain expressions of each of these variables in terms of the other. And when we say in terms of the other, one of the objectives will be to express this in terms of the Mach number. To do that, let us just eliminate certain variables. For example, we may eliminate d rho from equation number 1 which is the continuity equation and equation number 4 which is the equation of state. So, from equation 1 and 4, we can write that dP by P minus d rho by rho is plus dU by U plus dU by U minus dT by T equal to 0. Say, this is equation number 6. So, this is one equation. Also, you may write dU by U in terms of dm by m and dT by T. That is also possible and you may write dU by U in terms of dT by T. So, you may use equation number 3 and 4 to express dU by U in terms of the, sorry, 3 and 5 to express dU by U in terms of the other variables. So, let us use that 3 and 5. So, in the previous step, we eliminated d rho by rho. In this step, we will eliminate dU by U. So, we can write here, Cp dT by m square into gamma RT plus dU. So, plus dU by U, dU by U is 3 and 5. So, this is the equation number 6. dm by m plus half dT by T that is equal to 0. This is from equations 3 and 5 by eliminating dU by U. So, from this expression, it is possible to express dm by m as a sole function of dT by T involving the Mach number or dT by T as a sole function of dm by m as a function of Mach number. So, if we take dT by T common, we have Cp by gamma R m square, then plus half is equal to minus dm by m. So, this means that we have dT by T is equal to minus dm by m by Cp by gamma R m square plus half. Without going too much into the algebra, let us give it an equation number, equation number 7. Without going too much into the algebra, we may have a very important observation which we will note. See, the denominator of this expression is always positive because m square R gamma Cp, all these are positive. So, from this we may conclude that if dm by m is positive, then dT by T is negative. That means if m increases, then T will decrease and if m decreases, then T will increase. This is a very important observation that we get from here. And remember this observation till now whatever we have made is independent of whether it is having a friction or whether it is having no friction because till now we have not yet utilized the fact that it is a flow with friction. And if you observe very carefully, you will see that the only place where it has been utilized that it is a flow with friction is the term in the box mentioned in the equation number 2 that is there. Now when you have this dT by T from here, you may get other expressions also in terms of dm by m. For example, if you refer to equation 3 now, so equations 3 and 7, so using equation 7 in equation 3, you may get du by u as a function of dm and m. Once you get du by u as a function of dm by m, it is possible to get other parameters. For example, dp by p as a function of dm by m by referring to equation 6. So now if you use this in 6, you will get dp by p as function of m and dm. Similarly if you now substitute that in equation number 4 which is the equation of state, you can get d rho by rho as a function of dm by m. So that you substitute in 4 to get d rho by rho as function of m and dm. That means we have been able to express in principle dT by T, dp by p, du by u and d rho by rho as a function of dm by m. But that does not allow us to calculate the change in pressure, change in temperature, change in velocity along the length of the duct because we still do not know how m varies with x. So how will we know how m varies with x? For that we refer to the equation number 2 which is the momentum equation. So if we refer to the equation number 2, what we will get? So you may substitute that dp by rho, so both in terms of the mach number, then du by u as a function of mach number and tau wall by rho u square in terms of the friction factor which depends on the Reynolds number of flow. So we know that how friction factor is related to the Reynolds number and the surface roughness by the considerations of viscous flow as we discussed for flow through pipes and ducts in one of our previous chapters. So when the variations of all the variables expressed in terms of the mach number and its differential is substituted in equation number 2, it gives the governing equation for the variation of mach number and from that we may obtain a mach number as a function of x. Therefore, the relative dependencies of the variables, temperature, velocity, density factor all depend on the mach number and that nature of the dependence is not dependent on whether the flow has friction or not. But how mach number varies with x? Very much depends on the friction in the flow by virtue of equation number 2 and therefore that mach number as a function of x when it is substituted in different relationships, there will be difference in results for flow with friction and flow without friction because the mach number for a given x will be different for flow with friction and flow without friction. But in terms of the mach number, the dependences appear to be independent of friction but it is dependent on friction implicitly because mach number as a function of x is dependent on the extent of friction in the flow. So this is about that how we can get an estimate of the variation of the temperature with mach number, the pressure with mach number, the density with mach number and so on. Now next thing is about the directionality of the process in these types of ducts. So if you have such a duct, the processes which take place inside this duct, these processes are in general adiabatic but these processes are processes which are in general adiabatic with friction. So if you recall that if you want to calculate or if you want to identify the direction of a process, we have to see that what is the direction in which the net change in entropy of the system plus surrounding is positive. So if we calculate the change in entropy of the system, so ds is Cp dt by t minus r dp by p for an ideal gas. Or if we consider its form in the previous step, we have Tds is equal to dh minus dp by rho where substitution of the ideal gas equation has given rise to the simplified step. Now if you want to identify a step, a state where s is maximum say, say we want to get a maximum entropy state. So s is maximum that will imply that ds is equal to 0 that means dh is equal to dp by rho. Also from the energy equation, we have dh plus u du is equal to 0. Therefore in place of dh, we can write minus u du is equal to dp by rho. So this is why substituting the energy equation. Next we may substitute the d rho by rho in terms of du by u or du by u in terms of du by rho from equation number 1 by using the continuity equation. So in place of du, we may write minus u du by rho. So this is minus u in place of du minus u d rho by rho is equal to dp by rho. This is from equation number 1. So from this, what follows is that the rho gets cancelled out. So you get u square is equal to dp by rho. Remember that dp by rho is also equal to c square. The square is equal to c square of the sonic speed. And therefore at the maximum entropy condition, u square equal to c square that means m square equal to 1 or m equal to 1. So this is a very important observation that at the maximum entropy condition, you have the mach number equal to 1. You could express the maximum entropy condition in different ways. So if you want to evaluate actually what is the change in entropy as a function of temperature, you may you have to basically integrate the expression of ds. You have one dp by t, you have another dp by p. So it is important that you eliminate dp by p or write express dp by p in terms of the other parameter. So if you refer to equation number 6, equation number 6 will give you dp by p is equal to minus du by u plus dt by t. And du by u in terms of dt by t, you can get from equation number 3. So in place of du by u, we will be writing minus of cp dt by u square plus dt by t. This is from equation 3. Now it is possible to write again u square as m square into gamma rt but again m as an unknown will appear. So to keep it explicitly as a function of t only, the temperature as a variable without bringing mach number in the manipulation, what we will do is, we will just refer to the definition of the stagnation temperature. So remember that if you have h naught as the stagnation enthalpy, then you have h plus u square by 2 equal to h naught. In terms of cp, it becomes cp t plus u square by 2 is equal to cp t naught. So in place of u square you can write 2 into cp into t naught minus t. So that we may substitute here to get dp by p. So cp by u square will become 1 by 2 dt by t minus 2 by 2 by 2. So this is t0. So if you look at this equation, cp divided by u square will be 1 half into t0 minus t that minus sign adjusted, it will become t minus t0 plus dt by t. So let us call it as maybe equation number. What is the last equation that we have considered? Maybe we call it equation number 8. So if we substitute this equation number 8 into the change in entropy equation, let us see what we get out of that. So we get from the change in entropy equation that ds equal to cp dt by t minus r of dp by p. So minus r by 2 dt by t minus t0 minus r dt by t. Noting that cp minus r equal to cv, so this becomes cv into dt by t minus r by 2 dt by t minus t0. This is ds. So we may integrate it from a state say 1 to the given state. So we can write s minus s1 is equal to cv ln t by t1 minus r by 2 ln t minus t0 by t1 minus t0. So this means that we are able to write s minus s1 as a sole function of t, t0 and t1. So with this background, let us try to make up sketch of the temperature versus entropy diagram for this kind of a case. So if you make up plot, you will see that first of all we have already seen that there is a maximum in the s. So there will be a maximum in the s and the curve looks like this. So if we identify this point at which you have the maximum in the s, we know that the value of the Mach number here is equal to 1. That is what we have shown. Now clearly if you see we need to find the directionality of the processes for the curve which is in the upper part of m equal to 1 and for the curve which is in the lower part of m equal to 1. Remember when we say the total change in entropy, it is ds system plus ds surrounding. Here ds surrounding is 0 because it is an adiabatic process. Because it is an adiabatic process, you do not have any change in entropy because of heat transfer with the surroundings. So the change in entropy is only due to the change in entropy of the system. So when the change in entropy is taking place, it is taking place in a direction such that the entropy is increasing. So when the entropy is increasing, if you consider the upper curve, it is moving towards m equal to 1. When you consider the lower curve, it is also moving towards m equal to 1. Now what are the differences when you are having a process which is going towards the maximum entropy, which is the natural spontaneous way by which a process may thermodynamically take place. But there are 2 different paths by which one way reach that process for this case. For the part above, if you look into it very carefully, you will see that as it is moving towards m equal to 1, its temperature is decreasing. So for the upper part of the curve, since the temperature is decreasing, the Mach number should be increasing. Now refer to these conclusions that we had from equation number 7, the relationship between the change in Mach number and the change in temperature, which is this equation number 7. So from this equation number 7, which is identified here, we had earlier concluded that if you have a positive dm by m, you will have a negative dt by t. So if you have a negative dt by t, it must be a positive dm by m. That means along this path, the Mach number is increasing because the temperature is decreasing. So that means you have m less than 1 here. So if the Mach number has to go to 1 and if it has to increase along that path, that means along this path, Mach number must be less than 1. So it is going increasing and becoming equal to 1. If you consider the lower part of the curve, if you have the temperature increasing from the same logic, we can say that the Mach number is decreasing. So since the Mach number is decreasing, it must be m greater than 1 here. So it is the upper part is m less than 1. Its temperature is decreasing. That means the density is decreasing, the pressure is decreasing whereas the lower part m is greater than 1, the pressure is increasing, the density is increasing and temperature is increasing so on. So the upper part corresponds to a subsonic flow and the lower part corresponds to a supersonic flow and these parts go through a point of maximum entropy where m is equal to 1. So this line in the TS diagram is known as Fano line which is a very interesting line because it shows the locus of all points in the TS diagram corresponding to m greater than 1 and m less than 1 and you can see that if the initial state is subsonic that is m less than 1, the process tends to go towards a state where m is equal to 1 which is the maximum entropy condition whereas if m is greater than 1, from that also the process tends to go towards m equal to 1 which corresponds to the maximum entropy condition. So if you have a duct of constant area and if you have friction in the duct then there are 2 possibilities of the inlet flow say one is a subsonic flow another is a supersonic flow. So if you have a subsonic flow at the inlet then the Mach number will be going on increasing till it traverses the length that is necessary to achieve a Mach number equal to 1 and that depends on the friction in the flow because we have seen that the variation of Mach number depends on the wall shear stress because if we substitute all the variables pressure density velocity all these variables in terms of Mach number in the momentum equation then you can see that the momentum equation gives the Mach number variation in terms of the wall shear stress. Therefore the wall shear stress dictates strongly that how the Mach number will vary with x. So it requires a threshold x for the Mach number to become 1 from an initially subsonic state. If the length of the duct is less than that threshold length then obviously Mach number equal to 1 or the sonic condition will not be achieved. On the other hand if the length of the duct is greater than that threshold length for achieving Mach number equal to 1 for the given frictional condition then obviously what will happen then the flow will be choked at a critical length and that will give the maximum flow rate. On the other hand if the Mach number is greater than 1 at the inlet then if it has the threshold length for which the Mach number may become 1 then at certain length it will have a Mach number equal to 1 and from a supersonic to that critical or the sonic state it may come. If the length is short enough then there may be shock close to the exit of the channel or exit of the duct. If the length is very large then it may be seen that relative to the length of the duct the location of the shock is moving closer and closer to the inlet for an initially supersonic flow. So it depends on whether the flow is initially subsonic or supersonic to figure out that what will be the physical change in the flow as the compressible fluid is moving along the duct. For all these cases the variation in the Mach number is important and the variation of in the Mach number depends on the variations in the temperature density pressure and velocity these dependences in form are sort of independent of whether the flow has friction or not but implicitly depends on friction because how the Mach number itself varies is strongly dependent on the friction which is manifested through the wall shear stress. So from this discussion we may conclude certain interesting things. One is that this flow may be considered as a more general case of the isentropic flow. Here if you substitute the wall shear stress equal to 0 all the conclusions that you get should be corresponding to isentropic flow. That means if you substitute the wall shear stress equal to 0 all the expressions that you get here will be the expressions corresponding to the isentropic flow and the change in entropy in that case will turn out to be 0. So the non-zero change in entropy is the sole consequence of the friction in this particular flow. If you have heat transfer obviously the situation will be algebraically more complicated. How algebraically will it be different? Let us figure out that in which equation it will be different. So if you call our first equation was the continuity equation it does not sense whether there is heat transfer or not. So that equation will remain unchanged. Our second equation was the momentum equation there friction itself appeared and it is again insensitive to whether there is heat transfer or not. Our third equation was the energy equation and this is very much dependent on whether there is heat transfer or not. So if there is heat transfer what will happen? If there is heat transfer then in place of h plus u square by 2 equal to constant this will be replaced by an energy equation with heat transfer. So this will be like h i plus u i square by 2 plus q is equal to h e plus u square by 2 where i and e are the inlet and exit sections and this is the rate of heat transfer per unit mass flow rate. So this heat transfer term will be the only new term that will feature in the energy equation. For other equations like the equation of state that will also not be altered and the property relationship that gives the sonic speed that also will remain unaltered. So the similar analysis will be valid but it will be more complicated because of appearance of a new parameter in terms of the heat transfer and therefore one has to separately consider this heat transfer and it is possible to write another differential form of energy equation considering this heat transfer by taking a small control volume and by making an energy balance over the control volume just like the momentum balance of the control volume gives the momentum equation. Similarly the energy balance over the control volume with the heat possibility of heat transfer at the wall will give a new differential form of the energy equation which is different from the case with no heat transfer at the wall. So that will be the only change in terms of in principle. So you will still have these 5 equations with 5 unknowns. You will have a heat transfer at the wall which you may prescribe and therefore your energy equation will be different because it will now involve the heat transfer. So you may manipulate all the equations to come up with another temperature entropy diagram which will involve not only the frictional characteristics but also the heat transfer in the flow and that is a more general case than the case that we have considered here where there is no heat transfer and then you can get such lines in the TS diagram known as the Rayleigh line just like the Fano line that you get here. So more and more general cases they may be treated in a more and more general way. Remember that the treatment that we have discussed in this particular course that is only a one-dimensional treatment to give you the essential physics of the problem. In reality none of these problems are one-dimensional. These are all 2-dimensional or in a more general way 3-dimensional problems and one needs to solve the proper governing differential equations in 2 and 3 dimensions to get the flow field and the density field and the pressure field in the compressible fluid flow medium in a medium where the compressible flow is occurring. But here as a part of the scope of this particular chapter in this particular course we have only got restricted to one-dimensional flow with an understanding that how to write the basic equations in terms of approximate one-dimensional analysis and how to get the how to express the dependent variables in terms of the independent variables and how to figure out that what will be the permissible direction in which different processes may take place under those conditions typically for more general case with friction being present and with heat transfer being present. With this kind of a background what we believe is that in higher level courses if you come up or come across situations where you require the analysis of compressible flows in greater rigor and in greater details these type of physical understanding will be of good help for you to begin with typically when you consider the more general partial differential equations by which you may have your analysis and get the results not by such a simple one-dimensional form but through more rigorous and complicated exercise of solving partial differential equations even in such cases this type of one-dimensional treatment will give you a significant amount of physical insight on the variations of different parameters in the compressible flow. With this we conclude this lecture. Thank you very much.