 So, welcome to this lecture on rocket propulsion. In the last class, we had started discussing on nozzles. We are discussing quasi 1 D flows, which are essentially flow through variable area ducts. We have said that if the area variation is small, then the flow properties are going to vary only along the direction of variation of area. In that case, even though the flow is three dimensional, we can consider them to be one dimensional flows, which are called quasi one dimensional flows. So, for those cases, all the flow properties will be function of x only or the one dimension t x etcetera. Then, we started discussing the flow through a stream tube with variable area with area A 1 at section 1 and A 2 at section 2 and this is the stream tube. We considered the flow to be quasi 1 D. Let us say the properties at the inlet of the stream tube are P 1, T 1, U 1 etcetera and at the exit, the velocity is uniform, U 2 pressure is P 2, T temperature is T 2 etcetera. So, this is the problem we are solving last time. For that, we had made certain assumptions that the flow is steady, inviscid, adiabatic, then no body forces and potential energy negligible. We have made these assumptions. With these assumptions, then we derived the continuity equation, momentum equation and energy equation. So, the continuity equation was rho 1 U 1 A 1 equal to rho 2 U 2 A 2. We call this equation 1. Then, we derived the momentum equation, which was P 1 A 1 plus rho 1 U 1 square A 1 plus integral A 1 to A 2 P D A equal to P 2 A 2 plus rho 2 U 2 square A 2. We call this equation 2. We have said that this term here, the presence of this term, which is the integral from area A 1 to A 2 P D A, which is essentially the contribution of its pressure forces acting on this curved control surface, makes this equation non-algebraic. Then, we started discussing our energy equation. So, till the end of last class, we had derived the energy equation for this system, for this control volume. We have shown that the energy equation will be P 2 U 2 A 2 equal to rho 1 U 1 plus U 1 square by 2 minus U 1 A 1 plus rho 2 U 2 plus U 2 square by 2 U 2 A 2. We have proved up to this till the end of last lecture. Now, let us continue from here. Let us simplify this energy equation little more. So, from this energy equation, we can rewrite this as P 1 U 1 A 1 plus rho 1 U 1 A 1 times E 1 plus U 1 square by 2 equal to P 2 A 2 U 2 plus rho 2 U 2 A 2 times E 2 plus U 2 square by 2. We can write it like this. Now, what we do is, let us divide both sides by rho 1 U 1 A 1. We know that rho 1 U 1 A 1 is equal to rho 2 U 2 A 2. So, what we will do is, the left hand side we divide by rho 1 U 1 A 1, right hand side we divide by rho 2 U 2 A 2. While doing so, we will get P 1 U 1 A 1 by rho 1 U 1 A 1 plus E 1 plus U 1 square by 2 is equal to P 2 U 2 A 2 by rho 2 U 2 A 2 plus E 2 plus U 2 square by 2. We get this. Now, as we can see here, we can cancel this off. So, now what we are left with is P 1 upon rho 1 plus E 1 plus U 1 square by 2 is equal to P 2 upon rho 2 plus U E 2 plus U 2 square by 2. So, here P is pressure, rho is density, E is internal energy. Then from the thermodynamic definition, internal energy plus P by rho is enthalpy. So, therefore, this term is H 1 which is specific enthalpy at 1 plus U 1 square by 2 is equal to, this is H 2 plus U 2 square by 2. So, what we have is H 1 plus U 1 square equal to H 2 plus U 2 square. Let me call this equation 3. So, this is the energy equation. As you can see that area is not appearing anywhere, area has been eliminated completely. This is the energy equation for a steady adiabatic quasi 1 D flow. Now, if we take it further, we had not defined a specific location for 1 and 2. Still, we have shown that at 1 and 2, this relationship is valid. So, if I consider this as a property at any particular location 1, then this is the stagnation property. If I consider that the flow is brought to 0 velocity adiabatically, then this becomes the stagnation property H naught. So, what this is showing is that H 0 1 is equal to H 0 2. It essentially means that the stagnation enthalpy is constant everywhere in the flow field. This is of course, valid only with the assumptions that we had made. Therefore, this is something that we get for a quasi 1 D flow, that the stagnation enthalpy is constant everywhere in the flow field. So, this completes our discussion on the conservation equations in the control volume approach, integral form. We started with the integral form of conservation equations. In order to get this term here, integral form is not enough. So, now what we will do is, let us look at the differential form. Integral form can very easily handle our continuity equation and the energy equation. But one parameter remains here, which unless we specify a variation of a, we cannot solve for this. So, now let us look at the differential form of the conservation equations. So, let me clean this part and we took at the look at the same problem, but little differently. Next what we do is, we look at differential form of conservation equations. First of all we start with our integral forms, starting with the continuity equation. In continuity equation, what we have shown for this problem is that rho 1 u 1 a 1 equal to rho 2 u 2 a 2, that we have just shown. We had now as we have not specified in a specific location for a 1 and a 2. Therefore, this shows that anywhere, if I pick any location, then rho u a is constant. Then rho u a is constant. So, the integral form of the conservation equation for this problem says that rho u a is constant. We use this, that if rho u a is constant, if we differentiate this, d rho u a is going to be equal to 0. Let me call this equation 4, that d of rho u a is equal to 0. Next let us look at the momentum equation. We consider a small portion of the control volume that we had considered earlier. Let us say at this section, pressure is equal to p area is a, velocity is u, density is rho. At this section, pressure is increased little bit to say p plus d p, area has increased little bit by a plus d a. The velocity has changed little bit by amount d u and the density has changed little bit by amount d rho. Let us consider that the length of this section is d x and we have this pressure force is acting on this side. Now, we have in the previous case, since we are considering a quasi 1 d flow, we got the momentum equation in x direction. So, let us do this again here, that momentum equation we write in x direction. The expression that we got earlier from the control volume approach, we use that only for this control volume. So, in that case, if I use consider this as the control volume, we get p a plus rho u square a plus. Now, if the pressure acting on this side is p, area variation is d a, then the pressure force is acting from the side as we have seen in our integral form is p d a. So, this is p d a equal to the pressure, the term on this side, this is our section 1, this is our section 2. Now, we write this for the section 2. For section 2, we have pressure is p plus d p, area is a plus d a, density is rho plus d rho, velocity is u plus d u square a plus d a. So, this is the right hand side of the momentum equation, which was the values at section 2. Now, what we do is, let us expand this right hand side. When we expand it, we will get equal to p a plus p d a plus d p a plus d p a plus d p d a plus the expansion term you can write. I am not going into the details of that. Then, what we do is, this p d a term will cancel off as you can see from here and all the second order terms, we will drop because they are going to be small. So, dropping the second order term, we will get a simplified equation, which will be. So, here what we are doing is, first we are expanding, then dropping the second order term. So, dropping second order terms, second and higher orders of course, there will be a spare here also. So, second and higher order terms we are dropping, then we are redoing this calculation. We get a d p like here again this p a term will cancel off. Similarly, this rho u square a term will cancel off as you can expand, you will see. So, finally, what we will be left with is a d p plus a u square d rho plus rho u square d a plus 2 rho u a d u is equal to 0. Let me call this equation 5. Now, let us come back to this equation 4. That is our momentum equation in terms of in a differential form. Let us come back to the continuity equation, equation 4. From equation 4, we have d rho u a is equal to 0. So, now if I differentiate this, I get d u equal to 0. Let me multiply both sides by u. Then I will get u square a d rho plus rho u square d a plus rho u a d u equal to 0. Now, this term I will take to the right hand side. So, I will get u square a d rho plus rho u square d a equal to minus rho u a d u. Let me call this equation 6. We will further simplify it by putting this term now into this equation. As you can see here, we have this term 2 rho u a d u, which is also present in the right hand side of this. So, I can take this back and put into this equation and then we will get a. We had 2 of them, I will just replace 1 because there were 2 of this. I just replace 1 of them. Then after doing that, because it has a minus sign, you will see that this term and this term will cancel off. This term and this term will cancel off. So, it will be left with a d p plus rho u a d u equal to 0. So, a d p plus rho u a d u equal to 0. Now, area can be cancelled off. So, what we have is d p plus rho u d u equal to 0 or d p equal to minus rho u d u. Let me call this equation 7. This is a very important equation in fluid mechanics. This is called Euler's equation. This equation is valid with the assumptions that we had made. Whatever our assumptions that the flow is steady, inviscid, adiabatic, no body forces, potential energy negligible and quasi 1 d. With that, this equation we have derived is a very important fluid mechanics equation called Euler's equation. Now, this is the differential form of our momentum equation then. Next, let us get the differential form of the energy equation. So, coming back to the energy equation, we have proved that the energy equation can be written as H naught is equal to a constant and H naught is equal to H plus u square by 2. So, that is equal to constant for the quasi 1 d's flow we are studying. So, H plus u square by 2 equal to a constant. If I differentiate this, I will get d H plus u d u equal to 0. Let me call this equation 8. So, this is the differential form of my energy equation. So, therefore, equation 4 was the differential form of the continuity equation. Equation 7 is differential form of momentum equation and equation 8 is the differential form of the energy equation. Now, after doing that, let us try to combine these two and get an expression for the relationship between area and velocity because that is what primarily you would like to find out. So, the next after having done the derived the differential form of equations, next we look at the area velocity relations. The next topic is area velocity relationship. This tells us that when the area varies in the manner that we have considered, how the velocity is going to change? We see the velocity is changing from u 1 to u 2. How it is changing in between? That is what we try to find out. So, starting from our equation 4, which was the differential form of the continuity equation, the rho u a equal to 0. If I differentiate this and then divide both sides by rho u a, I will get u a d rho upon rho u a plus rho a d u upon rho u a plus rho u d a upon rho u a equal to 0. Then, if I cancel this, I get a simplified form that d rho by rho plus d u by u plus d a by a equal to 0. Let me call this equation 9, which is another form of differential form of continuity equation. Now, let us go back to Euler's equation or equation 7, which we have proved that d p equal to minus u d u. This was our Euler's equation. What we will do is, we will write it as d p by rho equal to d p by d rho d rho upon rho equal to minus u d u. There was a rho here. d p was equal to minus rho u d u. There was a rho here. d p was equal to minus rho u d u. So, we just take the rho to this side. Therefore, this is the form of Euler's equation. Now, we had assumed the flow to be adiabatic. We have also assumed its inviscid or frictionless. Therefore, it is reversible also. So, if it is adiabatic and reversible, it is isentropic. So, we have assumed the flow to be isentropic. Then, let us go to the isentropic relationships. Since the flow is isentropic for isentropic process, we know this relationship that the speed of sound is less than the speed is defined as d p d rho for an isentropic process. This is the definition of speed of sound. So, we use this now. If we define this like this, I take this and put it back into this equation. So, this term here is d p d rho. So, then this becomes a square d rho by rho equal to minus u d u. So, let me write it here. Therefore, a square d rho by rho equal to minus u d u. Now, this I can simplify it as d rho by rho equal to minus u by a square d u. This can be written as minus u square d u upon a square by u and by definition u by a is Mach number. So, this is equal to nothing but minus m square d u by u. Let me call this equation 10. So, now we have bought in the Mach number into picture or through using by using the isentropic relationship and considering the speed of sound. Now, we take this and put it back into this equation equation 9. So, you can see here we have this term d rho by rho. So, this d rho by rho term I replace by this. Then we get minus m square d u by u. So, this square d u by u plus d u by u plus d a by a equal to 0. This we are getting from 9 and 10. Now, if I simplify this I will get d a by a is equal to m square minus 1 d u by u let me call this equation 11. So, if I look at this equation once again this involves my continuity equation and momentum equation and the definition of speed of sound. Energy equation is not there, but energy equation we had already included when we derived the other equations. So, what we are seeing here now in this equation is a relationship between the area and velocity. So, this is called area velocity relationship. It is a very important relationship and tells us a lot of stuff. Now, let us look take a closer look at this area velocity relationship and see how we can infer from this. So, let us look at the significance of this relationship. First what is the area velocity relationship telling us how the area is related to velocity for a given Mach number. Mach number is also present here. Now, first let us consider a case when Mach number tends to 0. When Mach number tends to 0 this term here this term here tends to 0 which essentially means that the coming back to this term tends to 0. So, what we will have is u d a plus a d u equal to 0. So, u d a plus a d u equal to 0 which essentially means that d a plus a d u is equal to a u equal to 0 which means that a u is a constant. Area times velocity is a constant. So, when Mach number tends to 0 area times velocity is a constant. Now, from continuity equation we know that rho a u is a constant which is generally true for any Mach number that we have shown. For Mach number tending to 0 we are showing that a u is a constant. Therefore, this is a constant and this is a constant which means that density is constant which means that the flow is incompressible. So, what we have done here is starting from the general equations where we had not made anywhere the incompressible flow assumption. We started with the general equations we have shown here that when Mach number is more tends to 0 density is constant the flow is incompressible. So, here now we are proving that if the Mach number is low then density can be considered to be constant or the flow can be considered to be incompressible. So far it is made at a statement that if density is constant Mach number is low or other ways. So, when we define incompressible flow we say that Mach number is low. Now, we are proving it from the first principles. So, this is a big significance of this. Next, case 2 if Mach number is bounded between 0 and 1. So, if Mach number is between 0 and 1 what kind of flow do we have? We have subsonic flow by definition. So, for subsonic flow now what is happening? Looking back at this equation we have this term M square minus 1. So, if Mach number is bounded between 0 and 1 M square minus 1 is going to be less than 0. So, this term is going to be negative less than equal to 0 it is negative. If this term is negative then as d u increases or d u greater than 0, d u greater than 0 means velocity is increasing. So, this term is positive this is of course, positive this term is negative. So, positive negative is negative. So, therefore, d a is less than 0. So, this implies d a less than 0 which essentially means that for a subsonic flow if we have to accelerate the flow the area must reduce. So, we should have a converging passage right that is a nozzle because when we have to accelerate the flow we have the nozzle. On the other hand if d u is less than 0 that is we want to reduce the velocity. If the velocity reduces d u is less than 0, but M square minus 1 is also less than 0. So, there are 2 negatives. So, product of these 2 should be positive. So, therefore, if d u is less than 0 this implies d a must be greater than 0. So, what we can say is that for a subsonic flow once again if you have to reduce the velocity the area must be increased and reducing velocity should be a diffuser. So, in a subsonic diffuser the area must increase only then it follows the or the confirms to the conservation laws. So, therefore, we have shown here that for a subsonic flow if you have a nozzle the area must decrease if you have a diffuser the area must increase. Other way round I can say that for a subsonic flow area is decreasing d u must increase which the velocity should be increased. So, there must be an acceleration if area is reduced. On the other hand if area is increased like in the case of a diffuser d u must decrease which means rather d u must be negative. So, u must decrease. So, therefore, the velocity should decrease if you have a diverging area. So, for a subsonic flow converging area means a nozzle flow accelerates diverging area means a diffuser flow decelerates which again is proved from the area relationship that we have. On the other hand so let us look at the third possibility now where the Mach number is greater than 1 supersonic flow. So, third if Mach number is greater than 1 we have a diffuser supersonic flow. In that case again this term m square minus 1 is now greater than 0 because m square is greater than 1. So, m square 1 is minus 1 is greater than 0. In this case then if d A is less than 0 if d A is less than 0 is this term is negative this is positive. Therefore, this must be less than 0. So, implies d u less than 0 which means that for a supersonic flow if the area decreases velocity decreases. So, therefore, if Mach number is greater than 1 then the velocity this is u 1 u 2 u 2 is less than u 1. So, from the area relationship we have shown here now that if the flow is supersonic as the area decreases the velocity must decrease that means for a supersonic diffuser the area must be converging. On the other hand if d A is greater than 0 then once again this is positive this is positive. So, therefore, this must be positive. So, d u greater than 0 means d A greater than 0 means d u should also be greater than 0 that means the velocity should increase. So, for a supersonic flow once again if the area increases the velocity must increase. So, we have a supersonic flow coming here u 1 u 2 u 2 greater than u 1. So, once again for a supersonic flow the nozzle should be with increasing area. So, this is from the area rule what we have shown is that for a supersonic flow velocity decreases in a converging area velocity increases in a diverging area. So, this discussions are very very important in coming to D La Hall nozzle now. So, what we have if I can summarize what we have discussed today one more case is remaining which is the limiting case before I summarize I will go to the limiting case four. When Mach number equal to 1 what happens with Mach number equal to 1 that m square minus 1 is equal to 0 which means that from this equation d A equal to 0 d A by A equal to 0 which means that the area is either maximum or minimum. So, when the Mach number is equal to 1 the area is either a maximum or a minimum area. So, let me now summarize what we have discussed from the area velocity relationship. We have proved first when Mach number is small we have density equal to constant which means we have incompressible flow. We have proved that when Mach number is less than 1 subsonic flow then as area increases velocity decreases sorry velocity decreases. So, what we have proved velocity decreases as area decreases velocity increases. We have proved that if Mach number is greater than 1 for supersonic flow as area increases velocity increases as area decreases velocity decreases. And we have also proved that for a maximum Mach number equal to 1 area is either a maximum or a minimum. So, these are the things that we have proved from the area velocity relationship. So, what we have seen then that for Mach number is equal to 1 we have a maximum area or a minimum area. Now, let us find out which one is the realistic physical solution should we have a maximum area or a minimum area. So, let us consider various options 1, 2, 3 and 4. First let us look at a subsonic flow. So, incoming flow is less than 1 and it is going through a converging diverging area therefore, at this point there is a minimum area that is present. Now, let us say let us look at this flow. Since the flow is subsonic here the area is decreasing then according to our description so far the velocity will increase. So, the velocity increases in this direction it is possible then that since we have a minimum area here that the Mach number becomes equal to 1. So, it is possible that Mach number is equal to 1 at this area because we would like to have either a maximum or minimum for Mach number 1. So, Mach number becomes 1 here when we go beyond that now the flow is still accelerating and then here the area is diverging. So, therefore, as the area is diverging flow once it crosses Mach 1 becomes supersonic. So, here Mach number is greater than 1 and then flow will further accelerate. So, velocity continues to increase till we get at the exit a supersonic flow. Therefore, in this case the velocity is continuously increasing from the inlet to exit. So, there is no discontinuity in the process. So, therefore, this is physically possible that we have a subsonic flow we take it through a converging diverging area then it will accelerate from subsonic to supersonic without any problem still remain maintaining an isentropic nature. Next let us look at the same flow, but for a diverging converging nozzle. So, that here the area is maximum. So, once again the flow is coming from this side a subsonic flow as it enters this duct the area is increasing. Therefore, here the velocity will decrease. So, now it slows down as it slows down it is becoming slower. So, Mach number is not approaching 1 and if Mach number does not approach 1 we do not have this maxima corresponding to Mach 1 because according to our area rule at the maximum area Mach number should be equal to 1 which is it will not attain then on this side now. So, it becomes more subsonic till it comes here then in this side this subsonic flow it is accelerating. So, it will again accelerate here on this side and then reach some Mach number, but still remain Mach less than 1. So, therefore in this case the flow will first decelerate then accelerate it is neither it is not either a nozzle or a diffuser and it will not attain Mach 1 or beyond. Therefore, physically we cannot accelerate a subsonic flow to a supersonic flow if you go through a diverging converging nozzle. Next let us look at this possibility where Mach number is greater than 1. So, we have a subs supersonic flow coming in and we have a minimum area. So, since the incoming flow is supersonic and we have a diverging converging passage velocity will decrease. So, the velocity decreases. So, here velocity increases here velocity will decrease velocity decreases till it reaches this point where area is minimum as the velocity is decreasing Mach number is also decreasing. So, at the minimum area it can reach Mach 1. So, it is possible that we have Mach number equal to 1 here after this when we come to this side of the Mach 1 line now the flow has become subsonic and the area is increasing. So, subsonic flow increasing area velocity will further decrease. So, at the end we can get a subsonic flow. So, starting from a supersonic flow we can get a subsonic flow by going through a converging diverging area. So, and we have a minimum area where Mach number is equal to 1. So, this is also physically possible when we look at the fourth option that a supersonic flow entering an area is diverging converging area initially Mach number is greater than 1 and it is a diverging area. So, velocity increases. So, it becomes more supersonic. So, it will not reach Mach 1 till the maximum area. So, at the maximum area Mach number is not 1 it is actually much more than 1 then it comes to diverging area sorry converging area. So, it will slow down. So, the velocity will decrease here on this portion. So, it can come up come out still supersonic, but depending on the area some other value, but it will not become subsonic. So, bottom line is if I look at these two options a diverging converging section we are going through a maximum area, but it is neither a nozzle nor a diffuser because the direction of variation of velocity is not monotonic is in decreasing increasing or increasing decreasing. Whereas, when we go through a converging diverging area then for this case a subsonic flow going through a converging diverging area can be expanded to a supersonic flow. So, this is a supersonic nozzle a subs supersonic flow going through a converging diverging area can be slowed down to a subsonic flow. So, this is a subsonic diffuser supersonic diffuser. So, therefore, a converging diverging passage confirms to all the discussion that we had for the area velocity relationship and gives us a physically realizable solution. Therefore, for practical purposes for either a supersonic nozzle or a supersonic diffuser we need to have converging diverging passages. So, that and another point is that at the minimum area which is called the throat of the nozzle or diffuser the mach number is equal to 1. So, this minimum area is a physically realizable solution which is called the throat where mach number is equal to 1. So, therefore, we see that out of the four possible conditions two are feasible where we have converging diverging geometry. So, therefore, next what we will do is we have shown that from the area velocity relationship converging diverging is the feasible solution. Next we will look at how to get the actual parameter. So far it is all qualitative we talk about increase or decrease. Next now we will use an isentropic relationship for a variable area duct or a converging diverging nozzle and get the actual variables. So, we will stop here today.