 So, in this lecture and the next possibly couple of lectures, we will be looking at flow, compressible flow through a passage of finite but varying cross sectional area, which is representative of flow through nozzles, diffuses and blade passages in turbo machines. In whatever we have discussed so far, you would have noticed that the area of the stream tube never really appeared in any of the calculations or derivations. We just said that it was one-dimensional flow and so you can probably think of the area of the stream tube to be arbitrary. So, it never appears in any of the calculations but now because we have a passage with a finite and varying cross sectional area, the area of the stream tube is well known. So, for example, if you look at this nozzle, the stream tube for example, from inlet to outlet would just be something like this that would be a stream tube. So, now you can easily imagine that the cross sectional area has to appear in the governing equation. So, there may be there are likely to be modifications in the governing equation. So, let us take a look at this. Now, the problem, the main difficulty that arises in this case is that the flow is no longer strictly one-dimensional. So, if you look at this geometry and if for example, you sketch a few velocity vectors, let us say at any cross section, maybe at this cross section, if you sketch a few velocity vectors. So, along the central line, velocity vector would be horizontal. It will have only one component. Whereas, as you go up towards the, let me just draw it slightly longer. So, as you go up towards the wall, the velocity vectors would start to sort of turn like this and eventually, they become parallel to the wall. So, in the same way as you go like this, so the velocity vectors become parallel. So, if you look at the velocity vectors here, we can see that it is a two-dimensional flow, two components of velocity are non-zero. So, two components. So, strictly speaking, this is not a one-dimensional flow, but actually a two-dimensional flow. So, that is the main difficulty that arises now compared with what we have done so far. Now, the variation in the cross sectional area is in a direction normal to the main flow direction. So, the velocity component in the normal direction is non-zero and that is what we are seeing here. So, you can see that this has two components of velocity, one along the flow direction and one perpendicular to the flow direction. So, this becomes non-zero. However, what happens in most of the applications that involve these kinds of varying area passages is that the stream wise component of velocity is much, much larger than the normal component of velocity. So, if I denote the x component of velocity as Vx and the y component of velocity as Vy where this is x and this is y, then it turns out that usually Vx is much, much larger than Vy. Now, as the nozzle becomes longer and longer, of course, Vy becomes larger and larger. So, for example, in the case of a rocket nozzle where the area between area ratio between the exit and the throat could be of the order of 100 which is quite high, the normal component of velocity itself would be of the order of maybe couple of 100 meter per second. But the stream wise component of velocity in such cases is likely to be the order of 1.2 to 1.5 kilometer per second. So, that is much, much larger than the normal component. So, that is the most important observation that we make in these sorts of applications flow through varying passages, varying area passages, I am sorry. So, the normal component of velocity is small when compared with the axial component. So, as a first approximation, we ignore the normal, the normal component and simply treat this as approximately one-dimensional or quasi one-dimensional. So, this is not strictly one-dimensional, it is strictly two-dimensional, but we will neglect the y component of velocity and treat this as approximately one-dimensional. So, such flows are called quasi one-dimensional flows. So, let me just clean up this diagram a little bit. So, governing equations for the for this flow in differential form, they are identical to what we had before. The differential form of the governing equations are almost identical. You will notice that area appears in this in the continuity equation. So, almost identical to what we had before and we treat the flow as isentropic unless there is a normal shock somewhere in the divergent portion. So, we will see that a normal shock is likely to occur in the divergent portion of the nozzle. So, the flow is considered to be isentropic. So, if I call this station x and I call this y. So, for the normal shock, of course, we know that entropy increases across the normal shock. So, S y is greater than S x, but the flow itself is taken to be isentropic with entropy S x in this portion and isentropic following the normal shock and the entropy increases across the normal shock. So, that is a discontinuous increase in the entropy across the normal shock. Otherwise, the flow is considered to be isentropic. Now, if I apply the governing equation between two stations, let us use a slightly different color here. So, let us say I apply the governing equation between stations, let us say 1, let us call this 1 and let us call this 2. Then the continuity equation looks like this, which is fine, but the momentum equation, notice that the momentum equation is slightly more complicated than the differential form led us to believe. The differential form is same as what we had before, but the momentum equation in finite form looks this way, especially this term is very different. This comes from the fact that momentum balance now has to take into account the force exerted on the fluid by the walls of the nozzle. So, this is the stream tube that we are considering and we are applying the momentum equation across this stream tube. So, the force exerted by the nozzle surface on the fluid along this control volume has to be taken into account. So, that is what this term brings in. Energy equation looks the same as before, S2 is equal to S1. So, please make sure that you understand the change in the finite form of the governing equation versus the differential form of the governing equation. Now, the differential form of the continuity equation, you will recall, look like this, rho times V times A equal to 0. So, if you use product rule, expand and then divide by rho times V times A, you end up with this form for the continuity equation. And if I take a quantity like this d rho over rho, which is the first term here. So, d rho over rho may be written as d rho over dp times dp over d rho. And this quantity itself is equal to 1 over A square. And I can substitute for dp from my momentum equation like this dp equal to minus rho V dV. So, I end up with this expression for d rho over rho, where m is equal to V over A. And if I plug this back into my original equation, I end up with this expression dA over A equal to m square minus 1 times dV over V, which is a very, very important relationship in flows through passages with varying area. This is called the area velocity relation, because it involves the area obviously and the velocity. So, it is called the area velocity relation. So, basically what this says is the following. If I have, so we can imagine, let us say several situations. Let us say subsonic flow and let us say supersonic flow. So, if I look at passage, let us say converging passage in supersonic flow and subsonic flow, what is that? In this case, since dA is negative, the passage is converging, dA is negative. This tells me that dV is going to be, so if you look at this, dA is negative. And in the case of subsonic flow, this is also negative, which then tells me that dV has to be positive. So, V increases in this case, the velocity increases. Now, in the case of supersonic flow, dA remains to be negative, because it is a converging passage. But now, m square minus 1 term is positive because m is greater than 1, which means that dV has to be negative. So, supersonic flow actually decelerates in a converging passage. So, in the same manner, we can show that in case it is a diverging passage, so let us, in case it is a diverging passage, then dA is positive. And in the case of subsonic flow, this term is negative, which then means that dV has to be negative. So, a subsonic flow decelerates in a diverging passage and a supersonic flow, so this term becomes positive. So, supersonic flow must accelerate in a diverging passage. That is what we have shown here. So, this is a diverging passage, it is a converging passage, so area decreasing, converging passage, this is a diverging passage. So, subsonic flow in a diverging passage decelerates, subsonic flow in a converging passage accelerates and exact opposite for supersonic flow. This is true whether the fluid is calorically perfect or not, remember we have not used the calorically perfect assumption so far. So, this is true for air, true for steam and true for refrigerant. So, we have dealt with m less than 1 and m greater than 1. So, the question that arises next is what happens when m is equal to 1? So, if you look at this expression, so as m goes to 1, dA has to go to 0. We look at it that way, as m goes to 1, dA has to approach 0 for velocity to remain finite. So, as m goes to 1, dA goes to 0. So, as we allow m to go to 1, dA goes to 0. In fact, dA becomes equal to 0 when m becomes equal to 1 or wherever m becomes equal to 1, dA has to be 0. So, what this says is the following. Remember the flow is still isentropic, isentropic flow in a passage of varying cross-section. The sonic state can be attained only at a location where dA is equal to 0 that is another way of stating this. So, the location can either be a minimum or a maximum in the cross-sectional area. So, dA is 0 at this location or at this station and dA is 0 at this station also. Now, so we have two situations where dA is equal to 0 and let us examine this much more closely. Is there any more that we need to infer from this relationship? So, let us say that m is equal to 1 in this case. If the incoming flow is a subsonic flow, then it will accelerate in a converging passage which is this one here and it will be able to accelerate to m equal to 1 and then the flow will decelerate in the diverging passage which is this one here and it will be able to leave without any difficulty. So, m equal to 1 is possible in this case at the throat. Let us say it is a supersonic flow. Now, of course, in the case of the subsonic flow as I said, it will accelerate here reach m equal to 1 and then depending on the exit condition, it can continue to accelerate and reach supersonic speeds or it will decelerate and remain subsonic. Both exit conditions are possible, subsonic exit condition and supersonic exit conditions are possible which one is realized in actual practice depends upon the exit conditions that are being maintained here. What is the exit pressure that is being maintained at the exit of the nozzle? Now, in the case of supersonic flow, so supersonic flow decelerates in a converging passage which means that it can decelerate and reach m equal to 1 and then supersonic flow actually as you can see here, so it reaches m equal to 1. So, from here onwards again depending on the exit condition, we can either have flow which after reaching m equal to 1, it can decelerate and leave at a subsonic Mach number that is possible or depending on the exit condition, it can accelerate slightly and also leave, it can accelerate and also leave as a supersonic flow at the exit. So, both both exit conditions are possible. But what this says is, you know the m equal to 1 state can be reached by a subsonic flow at entry to the nozzle or by a supersonic flow at entry to the nozzle. That is what this discussion suggests. Now, let us go to this case again, if you assume m equal to 1 in this case, let us say we assume m equal to 1. Let me write this in a different color. So, we assume m to be equal to 1 here. Now, let us say that the incoming flow is subsonic. So, subsonic flow decelerates in a divergent passage like this, which means that it cannot reach m equal to 1. So, it is already subsonic and it decelerates further, which means it cannot attain m equal to 1 at this location here. Now, let us assume the incoming flow to be supersonic. Supersonic flow accelerates in this divergent passage. So, a supersonic flow for which the Mach number is already greater than 1 will accelerate even further to higher values of Mach number. So, it cannot attain m equal to 1 at this location. So, which tells us that when m equal to 1 state or the sonic state is attained, it is always attained at a throat and not at a location like this. This when m reaches 1 at the throat, the condition is known as choking. We will discuss this in a minute. So, the flow is said to be choked and what is important is choking if it occurs can occur only in a geometric throat. That is what we just demonstrated that m equal to 1 is possible only at a throat. Now, please bear in mind that the sonic state if it occurs, it must occur at a geometric throat, but flow need not always choke at a throat. That is very, very important. If m is equal to 1, it must always be at a location where dA is equal to 0 for the velocity to be finite. So, this is a very subtle point and you need to understand this. So, if m is equal to 1 that is on the right hand side, then dA must be 0 for the velocity to be finite. So, if m equal to 1 occurs, then it will always occur at a point at a throat where dA is equal to 0. On the other hand, if dA is 0, then m need not be equal to 1. m can be any value it wants, but dV is equal to 0. dV can be equal to 0. So, if dA is 0, then m need not be 1 there. So, what this means is if m equal to 1 occurs is realized because of the conditions that we are maintaining in the flow, the downstream pressure, upstream stagnation pressure and so on. If the conditions are such that the sonic state is realized, it will always be realized only at the throat. However, the Mach number need not always be 1 at the throat, it can be anything else, again depending on the conditions that are being maintained. So, the so the counterpart or the converse is not always true. If the flow is choked, it will always choke at the throat, but just because there is a throat does not mean that the flow has to be choked there. We will understand or we will give an explanation for why this is called choking as we go along. But for now, we will use the term without paying too much attention to it. Now, we will derive a relation called area Mach number relation. Notice that the previous one was area velocity relation. Now, we will derive something called an area Mach number relation for choked flow of a calorically perfect gas. So, here we are going to make use of the calorically perfect assumption. Whatever we have said so far is valid for calorically perfect as well as fluids which are not calorically perfect. So, let us say that you know we have an also like this. So, the mass flow rate at any section can be related to the mass flow rate at the sonic state. Now, again remember we said that sonic state if it occurs it occurs at the throat, but sonic state need not occur at the throat. So, in this case let us say the sonic state does not occur at the throat. What we would then say as I mentioned in our discussion of the sonic state, we can always extend this to a we can extend this. So, this is an imaginary extension and we would say that the sonic state occurs at this location because it is a location of it is a throat. We can add a convergent section like this and it is a throat. So, the sonic state occurs there or alternatively if the conditions are such that sonic state can be attained here then we can have a sonic state here. So, regardless of whether this sonic state occurs here or here we may write this because it is a same stream tube mass is conserved in this stream tube. So, I may write m dot equal to rho times v times a at this at this station equal to rho star v star times a star either here or there does not matter. So, if you rearrange you can get this expression. So, rho naught over rho may be written in terms of Mach number like this and rho naught over rho star may be written again like this after setting m equal to 1 in this expression. Remember v star is the velocity of the fluid at the throat and since it is a sonic state v star is equal to a. So, if you substitute all these into the equation that we have for a over a star we get this. Now, t star over t may also be written in terms of Mach number like this t star over t is equal to t star over t naught times t naught over t. So, you finally end up with an expression that looks like this a over a star is equal to this which is. So, the right hand side is a function only of Mach number. This is called the area Mach number relationship. So, given the area of the passage the cross sectional location and the value for a star we can determine the Mach number at that location using this relation. Actually this equation gives two values for a given a over a star. For example, if you look at this nozzle let us say that the flow is choked at the throat. Then you can easily see I am sorry let us go here. Let us say that the flow is choked at the throat. Then you can easily see that the Mach number I mean the value for a over a star is the same at these two stations. So, at this station and at this station. However, depending on the pressure if the pressure in the downstream of the nozzle is maintained in such a way that the flow becomes supersonic here and you can see that for the same value of a over a star. Remember here a star is equal to a throat because the flow is choked at the throat. So, for the same value of a star you can see that the flow is subsonic here reaches m equal to 1 here and the flow is supersonic in the divergent portion. So, for the same value of a over a star you can get either a subsonic solution or a supersonic solution and we use whichever one is appropriate. Now, very important a star is equal to the throat area only when the flow is choked. Otherwise as we just showed here a star would be equal to this the throat area of this imaginary extension. If the flow is choked then a star is equal to a throat. Now this function a over a star is listed I am sorry this quantity a over a star is listed as a function of Mach number in the gas table. So, you may use that let us just quickly take a look. So, you can see that not only t naught over t p naught over p rho naught over rho or listed a over a star is also listed. So, both the subsonic solution as well as the supersonic solution are listed for a over a star. Now, we may write the mass flow rate at any section like this m dot equal to rho v a and if I rearrange this by using stagnation quantities as well as the Mach number we may write it like this. So, we basically write this as rho over rho naught and this v is replaced in favor of the Mach number and a is written as a over a star times a star. Remember we are saying choked flow of a calorically perfect gas which means a star is equal to a throat. So, this a over a star may be replaced with the area Mach number relationship. So, if you do that you finally end up with an expression that looks like this. So, m dot equal to p naught a throat divided by square root of t naught times this quantity in this in the inside the square root I am sorry. Notice that this quantity depends only on the nature of the gas. So, or depends on the molecular weight of the gas gamma depends on whether it is diatomic or monatomic. So, a throat is a geometric parameter. So, p 0 and t 0 are the stagnation pressure and stagnation temperature respectively at the inlet to the nozzle that is very very important. This is the stagnation pressure and temperature at the inlet to the nozzle. So, basically we are looking at a situation like this. So, flow comes in at a stagnation pressure of p 0 and t 0 presumably at a subsonic Mach number that is a most common scenario reaches m equal to accelerates to m equal to 1 on the throat and then depending on the exit condition either accelerates or decelerates. We will discuss that in more detail as we go along. So, the mass flow rate at any section in the nozzle whether we take a section whether we take a section here section here or section here or here mass flow rate at any section in the nozzle is equal to this. This is an extremely important expression in flow through nozzles and it determines the operation of not only nozzles, but other components that occur I mean that are encountered in real life where we have multiple components each of which can choke independent of each other. So, this is a very very important expression in gas dynamics. The most striking feature of this expression is that it does not involve any downstream quantity. So, as I said you know the quantity inside the square root depends only on the gas and the quantities this is a geometric quantity and these two p 0 and t 0 on stagnation conditions upstream of the nozzle. So, the downstream conditions are not seen in this expression at all, but yet we are saying that depending on the downstream conditions that we are maintaining the nozzle may be choked or not, but once the nozzle is choked what this is telling is downstream conditions may determine whether the nozzle is choked or not, but once it is choked the mass flow rate is no longer dependent on the downstream condition. It is a very very powerful inference from this expression. So, which means that if I try to adjust the flow through the nozzle by let us say having a controlled downstream of the nozzle I can adjust the flow only as long as the flow is not choked. Once the flow is choked the mass flow rate becomes a constant and it is no longer dependent on the downstream pressure or downstream pressure is sufficient. It is no longer dependent on the downstream condition. The only way to change the mass flow rate through the nozzle in this case is to change p 0 or t 0 without changing the nozzle itself. If you change the throat area then you are changing the nozzle itself without changing the nozzle if you want to increase the mass flow rate for instance then I must increase either p 0 or decrease t 0. This is a very very important inference from this expression. So, this we have already already said. So, once the flow is choked the mass flow rate that can be realized through the passage is dependent only on the upstream stagnation pressure, stagnation temperature and the throat area. This means that the mass flow rate cannot be controlled anymore from downstream that is by adjusting the exit condition. But of course, you cannot adjust the downstream condition so much that the nozzle unchokes. If it unchokes then of course, this expression is no longer applicable. This expression is applicable only when the nozzle is choked and because we assume a star equal to a throat. So, if you change the exit condition so much that the nozzle unchokes then this expression itself is no longer valid. So, that is something that you must keep in mind. In other words, this is the maximum mass flow rate that can be achieved by adjusting the back pressure. This is very important. This is not the maximum mass flow rate that the nozzle can swallow. This is the maximum mass flow rate that the nozzle can swallow by adjusting the back pressure. If I want to increase the mass flow rate beyond this I can simply increase p 0. So, you must bear that in mind. The mass flow rate can be changed at will by an adjustment of the upstream stagnation conditions or the throat area. You can change the throat area or the upstream stagnation conditions. Now, you may be wondering when we would ever want to change the throat area. This actually is a very viable and widely used controlled strategy in case of supersonic inlet, supersonic diffuses or supersonic inlets which are present in the engine of aircraft that fly at supersonic speeds. So, there having a variable throat area diffuses is actually very useful for controlling the flow in the diffuses for choking and unchoking the flow at will. So, variable throat area nozzles are actually available and are quite extensively used in real life applications in aerospace propulsion. Now, the inferences that we have drawn here are applicable whether the substance is calorically perfect or not meaning it can be steam or refrigerant. But the exact expression may not be applicable because we have invoked the calorically perfect assumption. So, whether it is air or steam or refrigerant there is a maximum mass flow rate that can be achieved by controlling the downstream pressure. So, once the nozzle chokes the mass flow rate cannot be changed anymore by changing the downstream condition. And however, the mass flow rate can be changed at will by changing the upstream stagnation pressure or stagnation temperature. Because you must bear in mind that since we are not using or since the calorically perfect assumption cannot be used in conjunction with steam or refrigerant, the notion of a stagnation temperature itself is not meaningful in that case. Stagnation temperature is replaced by the stagnation enthalpy. So, the expression will not be valid for substances which are not calorically perfect, but the inferences are all correct. So, what we will do in the next lecture is to take a closer look at the flow inside nozzle. We will look at two situations convergent nozzle and a convergent divergent nozzle.