 So, we can look at problem number 1 in compressible flow which is C f 1. So, what is given here is that there is a sound wave front across which the pressure rise is 30 Pascal and this sound wave front moves into stationary air which is at temperature 300 Kelvin and pressure of 101 kilo Pascal. And you have to calculate the or estimate the velocity induced in air and the temperature change across this way. So, let me just draw a very simple sketch to show what is given. We have this sound wave front which is moving with the speed A. The pressure is 101 Pascal, kilo Pascal rather ahead of the wave and the temperature is 300 Kelvin ahead of the wave and across the wave delta P is given as 30 Pascal. There is a small increase in pressure as we have seen earlier in the lecture delta P that is given equal to 30 Pascal. And what we have to find out is what is the velocity that is induced here. So, velocity induced and also the temperature change across the wave. These are the two things that are required to be found. So, one way of doing this is as follows. So, A the speed of this sound wave can be found out using square root of gamma RT which is square root of 1.4 R for air is 287 and the temperature is given as 300 Kelvin rather. So, once this is known you can find out the speed with which this sound wave is moving. So, that is first part and if you recall we had A squared equal to delta P divided by delta rho. This was our expression that we had obtained earlier. Now, this delta P is given as 30 Pascal and this A square is something that we can find as above. And therefore, the first thing that you can determine is change in the density delta rho. That is first thing that you can do and then use the mass balance equation which will give you delta V equal to A times delta rho divided by rho and this rho is to be calculated as P over RT and that pressure is 101 kilo Pascal R is again 287 and P is 300. So, this is the density rho of the air into which this sound wave is moving. This is and now you have everything delta rho is calculated up here rho is calculated using our ideal equation gas equation and the speed of sound is also calculated directly using the expression square root of gamma RT right here. And therefore, this will give you the velocity that is induced in air behind the sound wave. Now, the way to interpret this is the following that we have this sound wave which is moving with the speed of A here it is 0 velocity stagnant air and behind is what we induce this velocity given by delta V that is first part. The second part is to calculate the change in the temperature delta T across the wave. So, one way of doing this is you can use the energy equation which was the change in the enthalpy equal to A times the velocity induced and this is simply C P multiplied by delta T equal to A times delta V. So, if you know the specific heat of air you are done because both the speed of the wave and the velocity induced have already been calculated. This is one way of calculating it else you can use P equal to constant times rho raise to gamma between 1 and 2 where 1 is the condition ahead of the wave and 2 is the condition behind the wave. So, P 1 here is 101 kilo Pascal P 2 is given as 101 kPa plus 30 Pascal density you can calculate here using which you can calculate the density here and then once you have the calculated density here you can calculate using these two values the temperature here and knowing the temperature T 1 you can find out delta T equal to T 2 minus T 1. The reason I am pointing out this another way of doing this calculation is because if you perform it these in these two different manners you will see that there is a slight difference in the values that you obtain and I would like actually for you to think about why this little bit difference should exist between the calculations done in one manner and done in some other manner. So, please think about what could be the reason and we can discuss this order discuss that also in a later session if you want. So, this was the first problem I told in the lecture that you can solve up to problem C f 0.7 with the material that we have covered today in the lecture. So, let me go directly to C f 7. So, there are two parts to this problem again. So, here we are dealing with a shock wave across which the pressure ratio is 1.15. So, we have a shock wave with P 2 over P 1 equal to 1.15 and it moves down the duct into still air which is at a pressure of 50 kilo Pascal and 300 Kelvin. So, what we have is this shock wave moving. Now, let me use the symbol C for the speed of the shock it is not a sound wave remember and P 1 here is given as 50 kilo Pascal P 1 is 300 Kelvin and P 2 over P 1 is given directly 1.15. So, P 2 is simply 50 multiplied by 1.15 kilo Pascal and here the velocity is 0 ahead of the wave. What you are asked to find is the temperature and velocity behind the shock wave. So, P 2 and velocity. So, the first part is pretty straight forward the calculation of temperature. You can use the Rankine Eugonio relation which is basically a ratio of T 2 over T 1 as a function of P 2 over P 1 and gamma and the shock pressure ratio or the shock strength P 2 over P 1 is given as 1.15 and this gamma is for air 1.4. So, using this you can directly obtain this ratio T 2 over T 1 and the once you know the ratio T 2 over T 1 since T 1 is provided you will get temperature T 2. So, this will provide you T 2 since T 1 is given. So, that was it. For the second part what you can do is let me redraw this picture. So, we have the shock moving with the speed C. We have P 1 T 1 v equal to 0 P 2 T 2. Let me write it only this much right now. So, this picture is drawn in a lab coordinate system. What we mean by this is that we are basically sitting in this lab coordinate system and we are watching the shock wave move past us with a speed of C in still or stagnant air which is at conditions P 1 and T 1. So, usually what can be done to solve these problems is to analyze these kinds of situations from a shock coordinate system as what we had done in the analysis in the lecture which basically means that we want to make the shock steady, attach the coordinate system to the shock and from that coordinate system which is attached to the shock perform the analysis. So, to do that instead of this what we have is you basically add a component C in the opposite direction to everywhere then what will happen is this shock becomes steady and let there be some velocity v 2 earlier which now you can call as C minus v 2. This is the way you can analyze the problem and now since all conditions are given it is also given that the pressure ratio is 1.15 which means that the P 2 and T 2 values can be found. Once you analyze it in this fashion you can determine what is the velocity v 2 behind the shock. One other way of doing this is use the shock tables which have been provided to you. So, what can be done is that using the shock pressure ratio which is P 2 over P 1 you can read out what Mach number this P 2 over P 1 corresponds to and this will require most likely some interpolation within your shock table which is exactly similar as what you would do for example, if you are using the steam tables. So, the interpolation process is exactly the same and once you have determined what is the upstream Mach number now remember this upstream Mach number is with respect to the shock. So, once you have determined the upstream Mach number with respect to the shock here is what we have this is M 1 P 1 T 1 P 1 and T 1 known you can also calculate rho 1 if you want and once you know this M 1 using the normal shock tables you can immediately find out M 2. So, this is with the shock tables P 2 is anyway given to you T 2 is something that you found from part 1 of this problem rho 2 is something that you can obviously find out once you know P 2 and T 2 once you know all these quantities you can find out V 2 which is the velocity of the flow post shock or behind the shock, but with respect to the shock. So, with respect to the shock we will see this velocity V 2 here is what we have V 1 which is with respect to the shock. So, now once you have found the velocity with respect to the shock what you can do is you can remove this coordinate system from where it is attached to the shock and make it again the lab coordinate system. So, revert to the lab coordinate system and in doing so what you are basically doing is you are adding a component of C which is the speed of the shock into each of these velocity components to the right. So, what is this C? M 1 is simply V 1 divided by A 1 and C is simply M 1 times A 1 which is essentially the velocity V 1. So, using this then you can calculate in part 2 what is given is that this shock is not really moving in still air, but it is moving in air which has a bulk velocity of 100 meters per second which is toward the shock. The pressure ratio is still given as 1.15. Again in this case what you can do is you can analyze the problem from the shock coordinate system. So, what we have here is the shock to which the coordinate system is attached and what we will have is then 100 meters per second plus the speed of the shock itself and C minus the V induced P 1 T 1 which will give you rho 1 and P 2 equal to 1.15 times P 1 as the pressure increase across the shock. T 2 is again exactly calculated in the same manner as what you did earlier using the Rankine Eugonio relations and using this set then you can calculate V induced. So, the bottom line or one of the most important tip that you can utilize is that usually analyze the problem coordinate system which simply means that the shock is made stationary. You can use the normal shock tables in which case the entire problem can be solved in terms of an upstream Mach number M 1 with respect to the shock. With this the remaining problems between CF 1 and CF 7 that is the CF 2 to CF 6 you can try to solve the only sort of let us say involved problem amongst CF 2 to CF 6 is probably CF 2 which involves determining the speed of sound for a gas that obeys the Van der Waals equation of state and the important part in that first. So, in that second problem is the very first part which is requiring that you have to show that the dp d rho partial derivative of pressure with respect to density at constant entropy is equal to gamma times partial derivative of pressure with respect to density at constant temperature. So, if you are able to show that using the thermodynamic property relations that you have gone through last week. The second part then simply involves using the Van der Waals equation of state to calculate this dp d rho at constant temperature and simply multiplying that by gamma to obtain the speed of sound for a gas which obeys the Van der Waals equation. So, I think CF 2 is something that I suppose will take a little bit of effort, but CF 3 to CF 6 should be fairly straight forward once you go through what we have just discussed in the first problem and the seventh problem. Thank you.