 So, let us now move on to discussion of flow of steam through nozzles, okay. Now Dilavel in 1888 actually used a convergent divergent nozzle to accelerate steam to high speeds for use in impulse steam turbines. Impulse steam turbines typically use high velocity steam at the inlet, so used converging diverging nozzle in 1888 itself. The importance of studying the flow of steam arises from the fact that steam turbines are still largely used today for power generation. Much of the power generation today electricity generation is done using coal or nuclear power, so steam turbines are used, okay. The heat is used to convert water to steam and the steam then spins the turbines. So, steam turbines are very extensively used. So knowledge of flow of steam through such passages is important because it occurs in the turbomissionary blade passages in steam turbine and also in the nozzles that precede these blades. So the theory that we have developed so far especially the physical aspects of the theory are very general, okay. The accompanying choking of a nozzle, area Mach number relationship, I am sorry area velocity relationship and then choking in a nozzle and mass flow rate, not the expression but physics relating to mass flow rate of choked flow in a nozzle all carry over to steam and refrigerant when required. However, the expressions have to be looked at again because we had invoked the calorically perfect assumption a few times, okay. So wherever possible we will modify this but as I said before we would not be looking at aerodynamic shocks in steam, okay. Now the calorically perfect model you may recall used two important assumptions that the gas obeys the ideal gas equation of state PV equal to RT and internal energy is a linear function of temperature. In fact, this itself is a cascade of two approximations. So basically we said internal energy U which is normally a function of temperature pressure is approximated to be a function of temperature alone which is further approximated to be a linear function of temperature or rather CV times T. Now let us examine each of these assumptions in turn for steam. So this figure was shown earlier in the module on psychrometry and it was in fact it was discussed in detail in the previous course in the module on pure substances and properties of pure substances. It is a qualitative sketch where the grey area illustrates regions where steam behaves almost as an ideal gas or obeys the ideal gas equation of state to within 10 percent. Now one of the problems that we encounter and steam expands through a nozzle is that there is no guarantee that the initial state will lie in the ideal gas region. There is certainly no guarantee that it will lie in the ideal gas region at the end of the expansion because at the end of the expansion the steam the state is in the two phase region okay. So there is some doubt that it will be out that is it is unlikely that the steam the initial state of the steam will be in the ideal gas region. Now you may wonder how using the same figure we made the statement in psychrometry that we may essentially take the water vapor in the atmosphere to be an ideal gas which obeyed PV equal to or which obeys PV equal to RT. Now remember we said that we added one important suffix to that statement for the range of temperatures encountered in psychrometric application. Remember psychrometric application the pressure is more or less one atmosphere and the range of temperatures also was given right. So for that range of temperatures we may essentially assume the water vapor in the air to be which is superheated to obey ideal gas equation of state. Whereas for the sort of application that we have in mind now namely expansion of steam in nozzles or in steam turbine blade passages the range of conditions that we will encounter is quite vast and there is no guarantee that steam may be assumed to obey the ideal gas equation of state and you know certainly that the specific internal energy of steam you know from steam tables that the specific internal energy of steam is definitely a function of temperature and pressure okay. So both these assumptions are invalid in the case of steam at least for the range of conditions that we are looking at okay. So both these are invalid for the range of pressures and temperatures that we are looking at. So we have to abandon the calorically perfect model and move on to doing calculations with property tables okay. So there is no guarantee that state one will lie within the shaded region or will behave as a perfect gas. State two certainly will not lie in that region because the state at the end of the expansion is almost always in the two phase region. And again specific internal energy is a function of t, p. So even the thermally perfect assumption is not valid okay forget about the calorically perfect assumption even the thermally perfect assumption which is the first one here. This is thermally perfect this is also invalid. The only assumption that we will make here is that the steam does not become too wet because if the dryness fraction becomes too low then we have to use the theory of gas dynamics of two phase flows because the water droplets or the liquid droplets have higher amount of inertia that is usually slip between the gas phase and the liquid phase. In other words the liquid droplets will not be able to follow the gas phase quite faithfully. So we get into some non-equilibrium effects and also two phase effects. So we will assume that that is not the case here that the dryness fraction is sufficiently high. Now isentropic expansion of steam remember when we talked about quasi one dimensional flows particularly flow through nozzles we said the following when we discussed here as the working substance. So we said that the flow is isentropic throughout or in the case of a normal shock the flow is isentropic upstream of the normal shock wave and downstream of the normal shock wave. So we are basically looking at isentropic expansion in nozzles okay in this case in the case of steam without any normal shock so that is our focus here isentropic expansion of steam. Now isentropic expansion of steam may be written as PV raise to n equal to constant following PV raise to gamma equal to constant for a calorically perfect gas okay. Now the value of n itself has to be determined from experimental data and this was done long back and it was shown by calendar that if the steam is superheated then it essentially obeys PV raise to 1.3 equal to constant and if it is a saturated mixture at the beginning of expansion then the index n may be calculated using this expression. So the speed of sound or acoustic wave propagation speed may be evaluated in closed form in this case as square root of n times P times V. What is that for in case of air this would have come out to be square root of gamma times R times T and we know that air obeys ideal gas equation of state PV equal to RT okay. So the expressions are the same except for the fact that gamma is replaced by n in this case but not always you have to be very very careful when you do the calculations okay. So the momentum equation remains the same and it may be rewritten like this V DV equal to minus V times DP. The left hand side may be integrated because it is a perfect integral and that gives us V2 square minus V1 square over 2. The right hand side also may be integrated now since we know the relationship between P and the specific volume. So PV raise to n equal to constant. So the right hand side may be integrated between states 1 and 2 to yield an expression like this or we may write V2 equal to this expression. So basically what this says is starting at state 1 where P1 V1 and the velocity V1 are known. We may then go to state 2 and P2 V2 and other quantities are known may be evaluated. Now the energy equation looks like this same as before this may be integrated to give the following. Notice that we are not replacing H with CP times T. Generally in the case of steam nozzles there are any loss of generality. We may assume that the steam actually expands from a reservoir where the stagnation conditions P0 and V0 prevail and it is expanded to a final pressure of P. So we are taking steam at a reservoir condition of P0 and we are expanding it through a nozzle convergent or convergent divergent to a final pressure of P. If you do that then the velocity at the end of the expansion process will be given by this expression. V velocity at the beginning of expansion is 0 because we are starting from stagnation condition. This is isentropic expansion of steam from stagnation pressure of P0 to a final static pressure of P is given by this expression. So as P decreases the velocity V obviously will increase. So because we assumed that isentropic expansion of steam may be written as PV raised to n equal to constant we are able to derive closed form expressions like this. This may also be derived using the Mollier diagram or this may also be obtained using the Mollier diagram or the steam tables. I will demonstrate this in one of the work examples that follows. So let us say that we have a nozzle convergent or convergent divergent. So starting from stagnation condition let us say that we are expanding it to a pressure P. So the m dot at this section may be written as A times V over V and this itself may be replaced by using the expression for isentropic expansion of steam. So we may write m dot equal to this. Now we already have an expression for V from here when the steam is expanded from stagnation state to a pressure of static pressure P. So if you substitute that this is what we get for mass flow rate at this section where the pressure is P and the cross sectional area is A. Now for a given value of a cross sectional area A P0 and V0, as I keep lowering the pressure P, as I keep lowering the pressure P the mass flow rate increases up to a point. So for a fixed value of A P0 and V0 if I keep lowering P the mass flow rate through the nozzle keeps increasing up to a point when it reaches a maximum and it does not change beyond that. Notice that we are demonstrating choking in an alternative manner here. Now how do we evaluate that maximum mass flow rate? We can do that by differentiating this expression with respect to P and then setting the derivative to 0. So please keep in mind that we are keeping P0 constant, V0 constant and A also constant. So P0 is constant, this is constant, A is also constant, cross sectional A is constant. We are simply changing the downstream pressure to smaller and smaller value and we want to determine that value for pressure at which the mass flow rate becomes a maximum. So we differentiate this with respect to P and then if you set the derivative to 0 we obtain this expression which shows that for this value of pressure when the pressure P over P0 is equal to this the mass flow rate is a maximum. Corresponding to this P over P0 we may evaluate the velocity to be square root of NPV. So if you substitute this P over P0 into this expression for velocity we get after a lot of simplification we finally get this to be square root of NPV. So when the mass flow rate is a maximum remember we are pulling the flow through the nozzle we are keeping P0 constant and we are constantly lowering P which is the exit pressure. When we do that the mass flow rate reaches a maximum when the pressure exit pressure corresponds to this value and the exit velocity for this case is square root of NPV which we showed to be the speed of sound in steam. So we are basically proving whatever we have already shown for a calorically perfect gas. That when you pull the flow through the nozzle the mass flow rate is a maximum when the exit pressure is equal to P star and the exit velocity is equal to the speed of sound that is what we have shown here. So we may actually write P equal to P star for this case when the mass flow rate is a maximum. Notice that the analysis that we have done so far is very similar to what we did before we have extensively used PV raised to n equal to constant that is what has made us allowed us to derive these expressions in closed form. But notice that you know we really strictly speaking do not use a Mach number in this case we use velocity mass flow rate and so on but we do not really use or derive area Mach number relationships and so on. So what we will do in the next lecture is to go through a couple of examples which illustrate the concepts that I have just outlined.