 So, in this lecture, we will look at or we will work out a few examples involving flow of steam through a nozzle. The first example reads like this, steam at a pressure of 600 kPa and drainage fraction 0.9 in a steam chest expands through a conversion nozzle having a throat area of 0.002 meter square into a region maintained at A 400 kPa and B 300 kPa. In each case, determine the exit pressure, exit velocity and the mass flow rate. So, with the given information of P naught equal to 600 kPa and X equal to 0.9, we can get the specific volume at the stagnation state to be V0 equal to 0.2845 S0 and S0 equal to 6.2771 from the steam table. Now, X1 is given to be 0.9, so that means the index of expansion, so index of expansion is going to be 1.035 times, I am sorry, plus 0.1 times X1, so 1.035 plus 0.1 times 0.9 and this comes out to be 1.125. So, the index of expansion is 1.125 and P star may be evaluated using the expression that was given earlier, you may recall that P star is equal to 2 over n plus 1 raised to the power n over n minus 1. So, for this value of n, P star is equal to 347.7 kilopascal. Since, now for case A, since the given ambient pressure is greater than the sonic state, I mean the pressure corresponding to sonic state at the exit of the nozzle, the flow is not choked and the exit pressure P is the same as the ambient pressure. So, this is very important. We need to evaluate P star in the case of a convergent nozzle, definitely we need to evaluate P star, compare that with the ambient pressure and then determine whether the nozzle is choked or not. So, since the nozzle is not choked, the flow at the exit of the nozzle is subsonic, which means the exit pressure is equal to the ambient pressure. So, we may evaluate the specific volume and exit using this expression and the enthalpy drop between the stagnation chamber and the exit may be evaluated using the expression that we derived before like this, all the values on the right hand side are known. So, we get the velocity to be, velocity at the exit to be 368 meter per second and the mass flow rate through the nozzle may be evaluated since the area at the exit is given, mass flow rate will be evaluated using this expression m dot equal to Ae times Ve divided by specific volume at exit and it comes out to be 1.804 kg per second. Now, the same analysis can also be done in couple of different ways. One is using the molier chart, which I am not going to present, but the other one is to simply continue to use the steam tables. For example, we can do something like this. So, let us say that this is a TS diagram. So, the stagnation pressure is given to be 600 kPa and the dryness fraction is 0.9. So, this we can label as state 0. So, dryness fraction is 0.9. So, now this steam undergoes isentropic expansion. The final pressure alone is not known. So, once we know the final pressure, once we know the final pressure 347, I am sorry, 400 kilopascal, we need to determine the exit pressure. For that alone, we need to know the index of expansion. For all the other things, we can use either the expressions that we derived earlier or the steam table can be used. So, for determining the value of P star alone, at least in the case of convergent nozzle, we need to, even in the case of divergent nozzle also, convergent divergent also, we need to determine the value for P star using the expression that we derived before, this expression here that we derived before. For that alone, we require the value of N. So, once we make the determination that P is 400 kilopascal, the steam undergoes isentropic expansion to an exit state E like this, 400 kilopascal. The specific entropy S0 is known. So, Se equal to S0. So, using this value and the fact that the pressure is known, the dryness fraction at the exit may be evaluated. So, with these two pairs of properties P comma Se, we may determine Xe and He. So, from which we may evaluate H0 minus He and V can also be calculated, Ve may be calculated using the expression that is given here. So, for calculating P star alone, we require the value of N. Let us go to case B. In this case, the ambient pressure is less than the critical value. So, the region or the ambient pressure is 300 kilopascal, which is less than the critical value of 347.7 kilopascal. So, the nozzle is choked and the exit pressure Pe is equal to P star. So, using these expressions, we may evaluate the enthalpy drop from which we can calculate the exit velocity as 425 meter per second. The specific volume may also be evaluated in the same manner and the mass flow rate in this case comes out to be 1.84 kg per second. So, this is the maximum mass flow rate that the nozzle can pass provided we are adjusting the mass flow rate by changing the downstream pressure. Now, notice that since the nozzle is choked in this case, the exit velocity may be calculated either by using the enthalpy drop or by using the fact that the exit velocity is equal to the speed of sound. So, I may also use this expression square root of N times P exit times V exit to calculate the exit velocity which again comes out to be the same 425 meter per second. Alternatively, I may evaluate for this case also. So, let me just extend this. So, for this case also, so this pressure would become 347.7 kilopascal and the exit state would be equal to this. So, we follow the same procedure as before to evaluate. So, PE is known, SE is known, Xe may be evaluated, enthalpy at exit, enthalpy drop, then exit velocity. The advantage with this approach is that you do not have to remember any of these expressions. The only expression that you need to remember would be this one here, the expression for P star. So, apart from this, of course, N also has to be calculated but this is based on, this procedure is based on whatever you have learnt already in the first level course itself actually as a matter of fact. So, it would be relatively straightforward, you know, there is not too many things that are different from before. Of course, remember this expression itself comes from SFE, right? So, nothing new there also. But I leave it up to you to decide which method you want to use because the third method which I have not discussed here is the one that uses the Moliage chart. So, there you can do all these things. So, looking up, so locating the state in the Moliage chart is the same as what we do with the TS diagram. You can look up the value of Xe, enthalpy can be located, retrieved directly from the Moliage chart and so on. Once that is available, you can calculate the rest of the quantities. Let us go to the next example. Dry saturated steam enters a convergent nozzle at a static pressure of 500 kPa and is expanded to 300 kPa. So, the exit pressure is given. If the inlet and throat diameters are 0.05 and 0.025 meters respectively, determine the velocity at the inlet and the exit and the stagnation pressure. So, TS diagram for this case would look like this. This is 500 kPa, 300 kPa. So, dry saturated steam. So, this is state 1 is expanded. So, this is the exit state E. So, given that P1 equal to 500 kPa and the steam is saturated vapor, we may retrieve the following values V1, H1 and S1 from the steam table. So, the index of expansion may also be calculated using Zunus formula as 1.135. Now, in this particular problem, the question of determining P star does not arise because the exit pressure is given to us. So, we do not have to do that. So, if you proceed in this manner, then you can evaluate the exit specific volume using this expression. The enthalpy drop may be evaluated using this expression, which is equal to Ve square minus V1 square over 2 and mass flow rates at the inlet and exit are the same. So, we may relate V1 and Ve like this. So, if you plug this into this expression, we may evaluate exit velocity as 436.6 meter per second. And so, it follows that the inlet velocity is 69.6 meter per second. Notice that in this case, the problem statement, it is not given that the steam is expanded from a steam chest. In fact, we are asked to determine the inlet velocity as well as the inlet stagnation pressure. So, which means that the velocity at inlet to the convergent nozzle is finite and not equal to 0. Now, the speed of sound at the exit may be evaluated to be 447.4 using the expression square root of n times Pe times Ve. So, we may evaluate the stagnation enthalpy. Since the inlet velocity is known, inlet specific enthalpy is also known, we may evaluate the stagnation enthalpy like this. And S0 is also known because the stagnation process is an isentropic process. So, we go up like this. So, this is state 0. So, S0 is known, S0 is known with these two property values, we can get the stagnation pressure to be 507 kilopascal from the steam table. Now, for this value of P0, once we have P0 using the expression that we derived earlier, this one, we may evaluate P star to be 294 kilopascal. So, the given exit pressure is higher than this value confirming that the flow is not choked. The next example, and it is like this steam at 700 kPa, 250 degree Celsius in a steam chest expands through a nozzle to a final pressure of 100 kPa. As flow rate is 0.076 kg per second, determine if the nozzle is convergent or convergent divergent, the throat diameter, exit diameter, the dryness fraction at the exit. Assume the expansion process to be isentropic and in equilibrium throughout. So, since pressure and temperature are given, we know that the steam in the chest is superheated and we may retrieve these property values from the table. Since the steam is initially superheated, the index of expansion is 1.3 and the critical pressure in this case is 380 kilopascal. Since the steam is expanded to a final pressure of 100 kPa, this means that the nozzle has to be convergent divergent. Remember, when it is said that the steam is expanded to a final pressure, that means that is the pressure with which the steam leaves the nozzle, that is the exit pressure. So, the exit pressure is given. So, since the exit pressure is 100 kPa and the critical pressure is 380, that means that the fluid undergoes further expansion in the nozzle. So, if we were to sketch the situation, it would look something like this. So, the stagnation pressure here is 700 kPa. So, the critical pressure P star is equal to 380 kPa. Now, the steam is expanded to a final pressure of 100 kilopascal. So, that means that the further expansion to be possible in this case, the nozzle has to be a convergent divergent nozzle. So, that P is equal to 100 kPa. It cannot be expanded in a convergent nozzle beyond a pressure of 380 kPa. For the given stagnation pressure, it cannot be expanded in a convergent nozzle beyond 380 kPa. That is very important. In fact, that is something that we have said here also. So, in this case, again, it is given that the nozzle, the steam is expanded to a pressure of 300 kilopascal. Critical pressure is 294 kilopascal. So, it means that this means that the flow can be expanded in a convergent nozzle itself and the flow is not choked at the exit. The flow does not become choked until the exit pressure reaches this value for the given value of stagnation pressure. So, the same logic is applicable here as well. So, we infer from this that the nozzle is convergent divergent. So, with N equal to 1.3, we may evaluate V star equal to this. Now, this assumes that the flow is superheated at the throat. The steam is, I am sorry, the steam is superheated at the throat. So, we may evaluate V star and the velocity at the throat which is equal to the speed of sound and the throat area to be this from the given mass flow rate. So, the mass flow rate is given as 0.076 kg per second. So, from the given mass flow rate, we may evaluate the throat area and enhance the throat diameter like this. Now, you have to remember in this case, so the steam is initially superheated 700 kilopascal, state 0, 700 kilopascal and 250 degree Celsius. So, the steam now expands isentropically to a final pressure 100 kPa. This is the exit state. Expansion is isentropic. The sonic state is somewhere here that has been assumed and we have proceeded with our analysis based on that. So, this is the point where the process line crosses the saturated vapor line. Let us label this as G. Let me use a slightly different color for this. So, that is the point at which or pressure at which the process line crosses the saturated vapor line. And that the pressure there may be evaluated simply by using the fact that SG is equal to S0. And if you go to the table with this value of SG, you may you should be able to retrieve the fact that PG is equal to 212 kPa. So, the enthalpy drop from the stagnation state up to this state takes place in the superheated region. So, we may evaluate the enthalpy drop for this part of the process, the specific volume at G and then calculate the velocity at this point when it crosses into the two-phase region to be 701.14 meter per second. Now remember, we evaluated P star to be 380 kilopascal. And since PG is 212 kilopascal, our assumption that the flow is superheated at the throat is actually valid. So, this is 380. So, once the flow crosses the saturated vapor line, N is equal to 1.135. So, now we consider expansion from state G to state E. So, the enthalpy drop between HG and HE may be evaluated like this and the specific volume at exit may also be evaluated like this. But VE now has to be evaluated the slightly different expression. This we obtained from steady flow energy equation because the velocity is finite and actually quite high at state G. And so V exit may has to be evaluated like this and comes out to be 864 meter per second. And using the given mass flow rate, we may evaluate the exit diameter to be 13.5 mm. Now we know VE, we know PE. So, we may evaluate the dryness fraction at the exit to be 0.96 from the steam table. Or we can alternatively if you wish to use the steam tables for the analysis, you may proceed in the same manner as before. So, S0, S0 and V0 are all known from the given stagnation state. So, we can then exit pressure is given. So, with the given value, so SE is equal to S0, PE is also known. So, with these two property values, we may evaluate XE, HE and the specific volume VE. And then the exit velocity VE may be evaluated using the enthalpy drop from 0 to E. And we may also evaluate the exit diameter using the given mass flow rate. However, for evaluating the throat diameter, we need to use, we need to calculate the value for P star and then evaluate the same manner, the velocity and other properties there. So, you may use steam tables to do this calculation or you may also use this expression, both are equally acceptable. Only thing is for evaluating P star, you need to use the expression that we have derived before. So, in this case, if you do that, then with P star known and S star also known. Remember, S star is equal to, so P star is known, S star equal to S0 is also known. So, from this, we can then evaluate using these two properties. Since it is still in the superheated region, quantities such as H star and V star may be evaluated from the superheated tables. So, once H star is known, using the enthalpy drop from H0 to H star, V star may be evaluated. And once a specific volume is known and the mass flow rate is given, the throat diameter may also be evaluated. Let us see how this is done. So, from the given steam chest condition, we may evaluate H0 and S0. Pressure of the throat alone has to be evaluated using the expression that we gave before and after assuming that P star or the fluid, I am sorry, the steam is still superheated at the throat section. So, we get P star to be 3IT. So, S star is equal to S0 and with the pressure and S star, we can go to the table and retrieve V star and H star. So, the velocity at the throat may be evaluated from SFE using this expression and the throat diameter may be evaluated using the fact that V star is known. We have retrieved it from the steam table. V star has been retrieved from the steam table. So, the throat diameter may be evaluated. Now, exit pressure is known, I mean specific enthalpy at the exit is also known and the fluid at the exit is a saturated mixture. So, the dryness fraction Xe may be evaluated as 0.96 from which we can get He and Ve from the steam table. And so, we may now evaluate Ve based on the enthalpy drop between state 0 and state E and similarly for the exit diameter. So, I find that in class many students actually like to use the tables for doing the calculations because they are very comfortable using the tables and the only expression that you have to remember is the one for evaluating N and the one for evaluating P star. So, once you have to remember only those two expressions, once you know those two everything else can be done either using the table or using the molier chart. So, this concludes not only this module but also the course. I hope that you found the lectures to be useful and it improved your understanding hopefully to a great extent, at least you know to a small extent. And if you have any feedback, please feel free to communicate them to me through the NPTEL site. Thank you.