 Hello and welcome to lecture number 23 of this lecture series on turbomationary aerodynamics. We have been talking about axial turbines in the last several lectures and we have had a chance to discuss about different aspects of axial turbines starting from very fundamental thermodynamics of axial turbines and moving towards 2 dimensional cascade analysis and also the 3D analysis which we are going to do subsequently. So, based on what we have discussed so far on the basic thermodynamics and the 2 dimensional analysis as I mentioned in the last class it is time that we have a tutorial session on understanding of axial turbines and how to solve problems which are related to axial flow turbines. But, before I go into the tutorial let me just quickly recap what we had discussed in the last several lectures. We had an introductory lecture on axial turbines where we discussed about different types of axial turbines like different types of turbines in general like the axial the radial and the mixed flow turbines of which the axial turbine happens to be the one which is most commonly used in especially in aero engine applications and also in marine as well as land based power plant applications. For a variety of reasons basically to do with the efficiency and the convenience in arranging axial turbines as compared to the other counterparts like the radial or the mixed flow turbines. Now, when we talk about the axial turbines alone we can further classify axial turbines based on the nature of the flow through the turbine itself. So, based on this classification we could have either an impulse turbine or one could have reaction turbine and so the turbine axial turbine may operate in either of these modes either an impulse turbine wherein the entire pressure drop takes place just in the nozzle and there is no pressure drop taking place in the rotor. The rotor simply deflects the flow and that does not contribute in any way in the pressure drop that is taking place through the turbine. So, this is true for an impulse turbine and the reaction turbine is one where the pressure drop is actually shared between both the nozzle as well as the rotor. So, part of the pressure drop takes place in the nozzle and the remaining part of that takes place in the rotor and so we have defined what is known as degree of reaction and how we can calculate degree of reaction. We also had a session on understanding the velocity triangles which are applicable for an axial turbine and also these different types of axial turbine like the impulse or the reaction type of axial turbine. So, wherein we discussed about how we can construct a velocity triangle starting from the fundamental principles and then we discussed about the various losses which occur in axial turbine and how some of these losses can be quantified. One may have two dimensional losses or one may have 3D losses like the secondary flows and typically cage losses and so on. We had a rather detailed discussion on these losses in our discussion on axial compressors and I think I mentioned that one could simply extend the discussion which was applicable for axial compressors also for an axial turbine. So, which is why we restricted our discussion on losses to the bare minimum because it is already been covered. We also defined different mechanisms or methods of calculating efficiency for axial turbines. There are two commonly used efficiency definitions the total to static efficiency and total to total efficiency. Each of them are used primarily according to the applications for which the turbine is being used. For example, if it is a land based gas turbine engine which is not expected to produce any nozzle thrust, one would like to expand the flow to the maximum extent possible without too much of kinetic energy leaving the turbine. In this case we would define the total to static efficiency, we would want the flow to have or to reach the static conditions without much dynamic head as it leaves the turbine. Now, in a narrow engine kind of an application where the turbine is meant only to drive the compressor and some accessories, one would still want some kinetic energy available at the turbine exhaust which can be further expanded in a nozzle and therefore, in such applications we would define the total to total efficiency. We have discussed some of these topics in the last several lectures and we also of course, discussed about the performance characteristics of axial turbines very important. We have discussed how what its significance is in relation to the engine as a whole and the significance of matching of turbine with compressors as well as matching of turbine with the nozzle. So, these were some of the topics that we had discussed in the last few lectures and in today's lecture which is basically a tutorial session. What we are going to do is to try and solve a few problems and use our understanding of the working of axial turbines and put that to practice in terms of solving problems. So, what I have for you today is a set of four problems on axial turbines which I will solve for you and after this I would also give you a few exercise problems which I would like you to go ahead and solve based on our discussion today as well as our understanding of these concepts in the last few lectures. So, this is what we are going to discuss in today's class. We will basically be having a tutorial session on axial turbines. So, let us go to the first problem. So, the first problem statement is the following. It states that a single stage gas turbine operates at its design condition with an axial absolute flow at entry and exit from the stage. The absolute flow angle at the nozzle exit is 70 degrees and at the stage entry the total pressure and temperature are 311 kilo Pascal and 850 degree Celsius respectively. The exhaust static pressure is 100 kilo Pascal. The total to static efficiency is 0.87 and the mean blade speed is 500 meters per second. Assuming constant axial velocity through the stage determine part A the specific work done, part B the Mach number leaving the nozzle and part C the axial velocity, part D the total to total efficiency and last part is to find the stage reaction. So, this problem statement is to do with the single stage gas turbine which is operating under certain design conditions the inlet stagnation pressure and temperature are given. The exhaust static pressure is given the total to static efficiency and the mean blades speed are given. Based on this data we are required to calculate a variety of other parameters and solve this particular problem. Now, as we have done in the past the with reference to turbo machines the very starting point of solving a problem if you recall when we had discussed about compressors is to get the velocity triangles right. So, the first point to this is to draw the velocity triangles and that is the starting point of solving any such problems associated with turbo machines. So, let us construct a generic velocity triangle here we are not given whether it is a reaction turbine or a impulse turbine nothing is mentioned. So, let us construct a generic velocity triangle like what we had done a few lectures earlier on and then from there we will see what are the parameters which have been specified and what is it that we need to calculate to be able to solve this problem. So, let us construct a velocity triangle in general. So, this is a very generic velocity triangle which is well which is also true for a reaction turbine stage it is not really an impulse turbine as you can see it is indeed a reaction stage. So, station 1 denotes the nozzle entry station 2 denotes the nozzle exit or rotor entry station 3 denotes the rotor exit. So, these are the three parameters that have been of the station numbers that we have defined and then the flow enters the nozzle with a velocity of C 1 and it exits the absolute velocity of C 2. Relative velocity at the nozzle exit is V 2 which is the velocity with which the flow actually enters the rotor and the corresponding blade angles have also been marked here u is the mean blade speed the flow exits the rotor with the velocity of V 3 and an absolute velocity of C 3 with their corresponding angles of alpha 3 and beta 3. So, this is a typical velocity triangle of an axial turbine stage and what we will do is we will take a look at what are the parameters that we know at this stage. So, from this velocity triangle we basically know the blade speed that something which has been specified and we also know the inlet conditions and exit static pressure and we are required to calculate a variety of other parameters. So, let us begin with the part a that is to calculate the first parameter which is to basically calculate the specific work done. Now, the first parameter that we need to calculate is the specific work done and yes we have also been given the flow angle at the nozzle exit that is alpha 2 has been given to us. So, there are two parameters in this velocity triangle specified we have alpha 2 and the blades p u. So, these are parameters which we know. So, let us calculate the specific work done we know that the total to static efficiency is basically a function of the specific work done with reference to the static pressure at the exit and isentropic conditions and therefore, the specific work done is basically total to static efficiency into C p into T 0 1 1 minus p 3 where p 3 is the static pressure at the rotor exit divided by p 0 1 raise to gamma minus 1 by gamma. So, these are parameters out of this all these parameters are known to us we know the total to static efficiency it is given as 0.87, C p is known 1148, T 0 1 is 1123, p 3 by p 0 1 is 1 by 3.11 because inlet condition is given as 311 kilo Pascal. So, that is 1 bar by 3.11 raise to 0.248 which is basically gamma minus 1 by gamma 0.33 minus 1.33 minus 1 by 1.33. So, we get 0.248. So, the specific work done or work done by the turbine can be calculated and it comes out to be 276 kilo joules per kilogram. Now, at the nozzle exit the Mach number we the second part of the question is to calculate the Mach number at the nozzle exit. Now, Mach number as we know is defined as the ratio of absolute velocity to the speed of sound. So, M 2 is C 2 divided by gamma R T 2. So, here we need to calculate two things one is to calculate C 2. We also need to calculate the static temperature at the nozzle exit that is T 2. Now, velocity triangle because it is stated that the flow enters and leaves the stage axially we have C w 3 is equal to 0 and since we know that w t is a product of u times delta C w that is u into C w 2 minus C w 3 and since C w 3 is 0 we can calculate C w 2 which is w t divided by u and that is 276 into 10 raise to 3 divided by 500 that comes out to be 552 meters per second. So, let us understand this a little better it is mentioned here that the flow enters the assuming a constant axial velocity through the stage and that the flow leaves the stage in an axial direction. So, we have the turbine which is operating in design conditions with an axial absolute flow at entry and exit from the stage. So, this actually needs to be modified in the sense that C 1 is axial C 3 is also axial as per this question. So, this general velocity triangle which I had drawn should actually have been modified in the sense that alpha 1 should be 0 alpha 3 should also be 0 and which is why we get C w 3 as 0 because since C 3 is axial C w 3 is also 0. So, C 2 we know from the exit velocity triangle is C w 2 divided by sin alpha 2 let us see that once again C 2 is C w 2 which is this component divided by sin alpha 2 and that is 588 alpha 2 is given as 70 degrees. So, C 2 comes out to be 588 meters per second we know that T 2 static temperature is T 0 2 minus C 2 square by 2 C p T 0 2 and T 0 1 are same because there is no change in stagnation temperature in the rotor in the stator there is no work done in the stator and so stagnation temperature in the stator has to be unchanged. So, T 2 is static pressure at rotor rotor entry which is T 0 1 minus C 2 square by 2 C p C 2 we have just now calculated. So, we can calculate static temperature at the nozzle exit. So, once we calculate static temperature we can now calculate the Mach number because Mach number is simply the ratio of absolute velocity to the speed of sound. So, we have already calculated the absolute velocity that is 588 meters per second and the static temperature 973 Kelvin. So, M 2 is 588 divided by square root of 1.33 into 287 which is the gas constant multiplied by 973. So, the Mach number comes out to be 0.97. So, that is that solves the second part of the question. Third part of the question is to find the axial velocity. Axial velocity can be found from the inlet velocity triangle because C 2 is known alpha is known. So, C a is C 2 times cos alpha 2 and that is simply 200 meters per second. Now, the fourth part of the question is to find the total to total efficiency. I think during our discussion on the efficiencies I had derived an expression which relates the total to total efficiency and the total to static efficiency and we will simply make use of that equation to calculate the total to total efficiency. Total to static efficiency has been given as 0.87 and so let us calculate now the total to total efficiency. So, based on that equation if you recall and go back to that lecture you will find that this derivation was shown. So, this is 1 by eta t t which is the total to total efficiency. This is equal to 1 by eta t s minus C 3 square by 2 W t where W t is the specific work done. So, all these parameters we have already calculated. So, we get 1 by 0.87 which is total to static efficiency minus C 3 square. Well, C 3 is basically equal to C a because the flow exits the stage axially. So, C 3 and C a are the same and therefore, this is 200 square divided by twice of W t 2 into 276 into 10 raise to 3. So, this comes out to be 1 by eta t t is 1.0775. So, the inverse of that eta t t is 0.93. Total to total efficiency we have calculated is 0.93. If you compare this with the total to static efficiency it is 0.87. So, you can see that total to total efficiency is indeed greater than the total to static efficiency. This is a comment which I had made even during our discussion on efficiencies that in general the total to total efficiency is comes out to be higher than the total to static efficiency. Because of the very nature of the definition I think I had shown the T S diagram to demonstrate why this efficiency total to total efficiency has to be higher than total to static efficiency. So, if you look at the expansion process here which is something I had mentioned in the during our definition of efficiency. Total to total efficiency is defined based on the temperature stagnation temperature here with reference to the stagnation temperature at the exit isentropic. Whereas, total to static efficiency is defined based on the stagnation condition here with the static condition here and the corresponding isentropic parameters. So, you can see that this difference is always less than this other difference and therefore, it is inherent that total to total efficiency has to be higher than that of total to static efficiency. So, that is coming from the very basic definition and therefore, that should be indeed true. So, this is something as which you have also seen in this particular problem that if you calculate that this actually should come out to be true. Let us move on to the last part of the question which is to calculate the degree of reaction. Now, degree of reaction again we have defined and derived an expression for degree of reaction in the one of the earlier lectures. So, this is basically equal to 1 minus C A by 2 u into tan beta 3 minus tan beta 2. So, from the velocity triangle let us see what these parameters are how do you calculate tan beta 3. Let us go back to the velocity triangle here tan beta 3 is equal to this component divided by the axial velocity that is u divided by C A and similarly, tan beta 2 is equal to the component given by this that is C w 2 minus u divided by C A. So, from these two equations we can calculate tan beta 2 as well as tan beta 3 and therefore, what you can see here is that tan beta 2 beta 3 is u by C A tan beta 2 can be equal to tan alpha 2 minus u by C A. So, if you substitute both these expressions here we can simplify the expression for the degree of reaction and that is 1 minus 1 by 2 C A by u into tan alpha 2. So, all these parameters we have already calculated and so, let us just substitute that here and then we get the degree of reaction as 0.451. So, we have now calculated all the 5 parameters which were required to be calculated for this question. We have calculated this calculated specific work done axial velocity the total efficiency and now the degree of reaction as well. So, that completes the first question which was to do with a very simple single stage axial turbine with certain parameters which have been specified and how do you proceed towards calculating the other parameters and which obviously, started from the velocity triangles trying to solve the velocity triangles to calculate the flow angles and therefore, the other parameters which were required to be calculated. So, let us now move on to the next question. So, this is the problem statement for this question this is basically a 50 percent reaction stage question. So, the problem statement is the following combustion gases enter the first stage of a gas turbine at a stagnation temperature and pressure of 1200 Kelvin and 4 bar. The rotor blade tip diameter is 0.75 meter, the blade height is 0.12 meter and the shaft speed is 10500 rpm. At the mean radius the stage operates with the reaction of 50 percent a flow coefficient of 0.7 and a stage loading coefficient of 2.5. Determine part a the relative and absolute flow angles from the stage, part b the velocity at nozzle exit, part c the static temperature and pressure at nozzle exit assuming a nozzle efficiency of 0.96 and the mass flow. So, here this is a question which pertains to a 50 percent reaction stage and we have the temperature and pressure at the inlet the degree of reaction is given to us the rotational speed and also the mean diameter is specified. And based on this we are required to calculate basically solve the velocity triangle and calculate the flow angles the relative and absolute flow angles. Part b was to calculate velocity at nozzle exit that is the absolute velocity then static pressure and temperature at nozzle exit with a certain nozzle efficiency which has been specified and also the mass flow rate. So, these are the parameters that we will need to calculate for this particular question. So, as always we will start with the velocity triangle let us construct the velocity triangle first and then we will proceed towards solving this question. So, this is what we had drawn for 50 percent reaction stage during our discussion on this couple of lectures earlier on. So, this is a velocity triangle for a typical 50 percent reaction stage and the main feature that you can observe for a 50 percent reaction stage is that the velocity triangles are symmetrical or mirror images. So, what you have at the inlet of the rotor is what happens at the exit as well it is just a mirror image and which means that the angles should also be equal alpha 2 will be equal to beta 3 and beta 2 will be equal to alpha 3 and so on. Similarly the velocity components if the axial velocity is constant then we have C 2 is equal to V 3 and V 2 is equal to C 3. So, in this question we have been given the blade height at the mean diameter and the speed which means that we can calculate the blade speed that is u flow coefficient is given as 0.7 which means we can also calculate the axial velocity from there because flow coefficient is the ratio of the axial velocity to the blade speed. So, we have the blade speed as well as the axial velocity we are now required to solve the velocity triangle and calculate a few other parameters. So, let us start with the first part of it which is to calculate all the angles involved in the velocity triangle. Now for this case we also have been given the loading coefficient psi which is delta H naught by u square and for a 50 percent reaction stage delta H naught is equal to u times delta C w which is also the same as delta V w and therefore the stage loading coefficient reduces to delta H naught by u square which is V w 3 plus V w 2 divided by u. And so this if we express in terms of the angles let us take a look at what is V w 2 and V w 3 V w 2 corresponds to this component here which is in terms of beta 2 here and the axial velocity. Similarly, V w 3 is for the exit in terms of the axial component and the blade angle that is beta 3. So, this we have expressed in terms of axial component V w 3 is C a times tan beta 3 and V w 2 is C a times tan beta 2. So, from our discussion on degree of reaction we specifically for 50 percent degree of reaction case we can basically equate degree of reaction as C a by u into tan beta 3 minus tan beta 2 divided by 2 and that is because the angles are similar. So, if you replace alpha 3 by beta 2 and beta alpha 2 by beta 3 we can express degree of reaction in terms of this. So, if you look at these two equations and we try to simplify them what we basically get is an expression in terms of the degree of reaction the loading coefficient and the flow coefficient C a by u is the flow coefficient which we know is equal to 0.7 in this question. So, what we get is tan beta 3 is equal to psi by 2 plus degree of reaction divided by the flow coefficient phi. Similarly, tan beta 2 is equal to psi by 2 minus degree of reaction divided by phi. This basically happens if you simply add and subtract these two equations we can simplify them and get this equation. So, since we have been we already know the loading coefficient psi the flow coefficient C a by u and degree of reaction we substitute for those values here and we get the angles beta 3 which is 68.2 degrees and beta 2 which comes out to be 46.98 degrees. Now, so for at 50 percent reaction stage we do not need to actually calculate the other angles because other angles are equal to what we have already calculated alpha 2 will be equal to beta 3 which is in turn equal to 68.2 degrees and alpha 3 is equal to beta 2 which is 46.98 degrees. So, in this question that we have because it happens to be a 50 percent reaction stage we could make lot of simplifications in calculation of degree of reaction or equating that to the loading coefficient and so on. And also the fact that we do not need to really calculate the other angles because beta 2 is equal to alpha 3 and beta 3 is equal to alpha 2. So, we have just calculated beta 2 and beta 3 and the absolute angles are indeed equal to these angles as well. So, that completes the first part of the question. Now, let us look at what is the second part. Second part is calculate velocity at nozzle exit. Subsequently we need to calculate static pressure and temperature at nozzle exit with a certain nozzle efficiency specified. So, let us calculate the velocity at the nozzle exit and for which we will make use of the fact that the dimensions have been specified the mean the radius has been given the blade speed is known and so we can basically calculate the axial velocity and also the absolute velocity. Now, at the mean radius we can for which the tip diameter and the blade height will be used. So, the tip diameter is given as 75 centimeters. So, 0.75 minus the blade height 0.12 divided by 2. So, this is the mean radius that is 0.315. So, at the mean radius we can calculate the blade speed the speed of rotation is given here and the. So, pi d n by 60 will give us the blade speed mean blade speed that is 346.36 meters per second. Since, the flow coefficient has already been specified we can calculate the axial velocity as 5 times u mean that is 70 percent of this and so that is 242.45 meters per second. So, the absolute velocity at the nozzle exit is C a by cos alpha and alpha is given as we have already calculated in the previous part of the question that was 68.2 degrees. So, C 2 is equal to C a by cos alpha 2 and that is 652.86 meters per second. So, having calculated the absolute velocity we can now calculate the static temperature based on this absolute velocity because stagnation temperature is known it is given as 1200 Kelvin. So, in this question we have been given the stagnation temperature at the inlet of the turbine as 1200 Kelvin and we know that in the nozzle there is no change in stagnation temperature. So, T 0 1 should be equal to T 0 2 and so T 2 is basically equal to T 0 1 minus C 2 square by 2 C p and since that is because T 0 1 and T 0 2 are the same. So, from there you can calculate static temperature static temperature at the nozzle exit will be equal to T 0 2 which is T 0 1 minus C 2 square by 2 C p. So, we get 1200 minus C 2 square 652.86 square divided by 2 into C p. So, this can be calculated at 1016.3 Kelvin. Next part is to calculate the stagnation or static pressure at the nozzle exit for which we will make use of the nozzle efficiency definition. So, this is again where from the fundamental cycle analysis that I assume you would be aware of for Brayton cycles where we could where we have defined nozzle efficiency in terms of enthalpies at 0 1 being the inlet enthalpy of the nozzle H 2 is the inlet static exit static enthalpy divided by at 0 1 minus H 2 S. So, we can express this in terms of temperatures 1 minus T 2 by T 0 1 divided by 1 minus P 2 by P 0 1 raise to gamma minus 1 by gamma. This is because we have we can express the denominator in terms of 1 minus T 2 S by T 0 1 which is basically in terms of the pressure ratios for an isentropic condition. So, from this we can simplify this and calculate the pressure ratios P 2 by P 0 1 in terms of the temperature ratio as well as the nozzle efficiency. And so, the static pressure at the nozzle exit can be simply calculated in terms of inlet pressure stagnation that is 4 bar multiplied by 0.84052 raise to 4.03. So, this is 1.986 bar and the last part of the question is to calculate the mass flow rate. Mass flow rate is expressed in terms of density, the area and the axial velocity. Now, density is P 2 by R T 2 and how do you calculate the annulus area? So, for calculating this area we have the tip diameter as well as the blade height. So, from there you can calculate the annulus area and multiply that with the axial velocity one can calculate the mass flow rate. So, density is P 2 by R T 2 annulus area in terms of the blade tip diameter and the blade height and the axial velocity which is already something we have calculated. So, if you multiply all of them we get mass flow rate as 39.1 kilograms per second. So, this completes the second question which was very similar to what we had solved in the first question except for the fact that here we had also firstly a reaction turbine 50 percent reaction turbine stage. The first question was on an impulse turbine stage and the second difference being the fact that here we had nozzle efficiency and correspondingly we had calculated the static temperature and pressure at the exit of the nozzle using the nozzle efficiency definition. So, having understood two distinct problems one to do with an impulse turbine and this second question was a 50 percent reaction turbine we will now proceed to solving two other problems which are slightly different from what we have already solved. So, the third question the problem statement is the following the inlet a single stage axial flow turbine operates with an inlet temperature of 1100 Kelvin and a total temperature of 3.4 bar. The total temperature drop across the stage is 144 Kelvin and the isentropic efficiency of the turbine is 0.9. The beam blade speed is 298 meters per second and the mass flow rate is 18.75 kilograms per second. The turbine operates with a rotational speed of 12000 rpm. If the convergent nozzle is operating under choked condition determine part a the blade loading coefficient part b the pressure ratio of the stage and part c the flow angles. So, in this question we have been given that the nozzle is operating under choked condition that is one of the things that we need to keep in mind. We also have been given an isentropic efficiency for the turbine besides few other parameters like the temperature and pressure the temperature drop in the stage blade speed and the mass flow rate. So, as we have been doing in the past we will construct the velocity triangle for a generic turbine axial turbine and so that we are familiar with the parameters which are specified and those which we need to calculate. And so in this question we have the blade speed which is given 298 meters per second and the mass flow rate is given to us and the pressure and temperature at the inlet have been specified. So, let us solve the first part of the question which was to calculate the loading coefficient which is very straight forward because we know the temperature drop and we know the blade speed. So, psi is simply C p times delta T naught divided by u square. So, that comes out to be 1.8615. Now, second part of the question is to find the pressure ratio of the stage. So, we need to find P 0 1 by P 0 3. So, to calculate that we know the temperature drop in this stage it is given as 144 Kelvin. The inlet temperature is given T 0 2 is equal to T 0 1 that is 1100 Kelvin. Based on that we can calculate the exit stagnation temperature T 0 3 which will be T 0 1 minus delta T naught and that is 1100 minus 144. So, that is 956 Kelvin. So, having calculated the exit stagnation temperature we can we need to also calculate the pressure ratio because that is the objective of the second part of the question. For this we will make use of the nozzle efficiency definition. So, isentropic efficiency I mean the isentropic efficiency of a turbine which is defined in terms of the actual drop in stagnation temperature divided by the isentropic drop. So, isentropic efficiency is defined as 1 T 0 1 minus T 0 3 divided by T 0 1 minus T 0 3 S. Humorator is already known that is delta T naught divided by T 0 1 into 1 minus P 0 3 by P 0 1 raise to gamma minus 1 by gamma. So, if you simplify this because the efficiency is already given to us we can substitute for the efficiency the temperature drop is known inlet stagnation temperature is known and therefore, P 0 3 by P 0 1 comes out to be 0.533 and therefore, the pressure ratio of the turbine is an inverse of that that is 1.875. So, that solves the second part of the question where we were required to calculate the pressure ratio of this particular turbine stage a single stage axial turbine. The next part is to calculate the flow angles we need to calculate all the angles that are involved in the velocity triangle. So, here we are going to use the fact that the nozzle is operating under a choked condition. So, that is one of the informations that we have and that means that the Mach number at the nozzle exit is 1 and therefore, the absolute flow at the nozzle exit C 2 will be equal to square root of gamma R T 2. And similarly, that also fixes the temperature and pressure at the nozzle exit because the temperature and pressure would be the critical conditions and therefore, we can actually calculate static pressure and temperature from the isentropic relations because it is a operating under choked condition with Mach number equal to 1. So, once we calculate the absolute velocity we can also calculate some of the flow angles that are involved because the blade speed is known and therefore, we can proceed towards solving the velocity triangle and calculating all the angles. So, the nozzle since it is operating under choked condition Mach number is exit Mach number is unity. So, C 2 is equal to square root of gamma R T 2 and therefore, T 2 by T 0 2 by T 2 is equal to gamma plus 1 by 2 and that is 1.165. Since T 0 2 is equal to T 0 1 that is known the static temperature at nozzle exit is T 2 which is 944.2 Kelvin and therefore, the absolute velocity of the gases leaving the choked nozzle will be square root of gamma R T 2 that is 600.3 meters per second. And now, we can calculate the axial velocity because the blade speed is known the flow coefficient is known. So, u times C a C a is equal to u times phi that is 298 multiplied by 0.95 that is 283 meters per second and from the velocity triangle we can now calculate cos alpha 2 which is C a divided by C 2 that is 283 by 600 and therefore, alpha 2 is 62 degrees. We now need to calculate beta 2 which is the blade angle at the inlet of the rotor and for which we will make use of the velocity triangle again and that is from the velocity triangle you can see that this ratio u by C a is tan alpha 2 minus tan beta 2 which is the inverse of the flow coefficient. Since tan alpha 2 is known and phi is also known we can calculate tan beta 2 which is tan alpha 2 minus 1 by phi that is 0.828 or beta 2 is 39.6 degrees. Now, the third part of the question well wherein we are calculating all the angles we need to now calculate the angles at the rotor exit. Now, the specific work of the turbine which is W t is C p times delta t naught which can be expressed in terms of u into delta C w that is u into C a tan alpha 2 plus tan alpha 3. In which case we know the angles at the inlet tan alpha 2 is known u is known C a is known and delta t naught is also known. Therefore, tan alpha 3 is C p times delta t naught by u C a minus tan alpha 2. So, all these parameters are known to us and we simply substitute that and we get alpha 3 as 4.54 degrees. Similarly, u by C a is also equal to tan beta 3 minus tan alpha 3 and therefore tan beta 3 is equal to 1 by 5 plus tan alpha 3 and therefore, beta 3 is 48.54 degrees. So, we have calculated all the angles that are involved starting from the rotor entry alpha 2 and alpha beta 2 and rotor exit the alpha 3 and beta 3. So, you can see that to be able to solve in fact, all the three problems that we have now solved requires you to have a good understanding of the velocity triangles because that is where that basically the starting point of solving these questions it is necessary that you construct the velocity triangles clearly understanding the different components of velocities which have been specified and then use that information to proceed and solve the problem based on what is already given to given in the problem like in some questions you have the mean blade speed some questions may specify axial velocity or some angles and so on. So, in this question we have basically solved we have calculated the pressure ratio we have also calculated all the angles that are involved and in this case of course, it is also given to us that the nozzle is operating under choked condition. So, I have one more question for you in this is different from all the three questions we have solved in the sense that this question requires us to find the number of stages that are required for operating a certain turbine. So, it is a multi stage turbine how do you calculate the number of stages if you know the pressure drop across the entire turbine how can you estimate the number of stages that are involved. So, this is a problem statement is the following a multi stage axial turbine is to be designed with impulse stages and is to operate with an inlet pressure and temperature of 6 bar and 900 Kelvin and outlet pressure of 1 bar the isentropic efficiency of the turbine is 85 percent all the stages are to have nozzle angle outlet of 75 degrees and equal inlet and outlet rotor blade angles. The mean blade speed is 250 meters per second and the axial velocity is 150 meters per second and is constant across the turbine estimate the number of stages required for this turbine. So, this is an impulse turbine and so it is needless to say that in the rotor inlet and outlet the angles are going to be equal it is also given that we have the inlet pressure and temperature for the nozzle entry we have been given the isentropic efficiency and the nozzle exit angle we also have the mean blade speed and the axial velocity. So, lot of parameters that are involved in the velocity triangles are given to us, but we need to calculate the number of stages that are required for getting this kind of pressure ratio that is involved in this particular turbine. So, let us take a quick look at the velocity triangle because we will need to come back to this little later. An impulse turbine would involve a velocity triangle like what is shown here C 2 is the flow exiting the nozzle at an angle of alpha 2 in this case it is given to us V 2 is the relative velocity entering the rotor at an angle of beta 2 and it leaves the rotor at velocity V 3 which is equal to V 2 and an angle beta 3 which is again equal to beta 2. The absolute velocities that the rotor exit are is equal to C 3 which is in fact equal to V 3 and V 2 for constant axial velocity. So, in this case it is indeed specified that the axial velocity is a constant. So, it means that C 3 is equal to V 3 which is in turn equal to V 2. So, we will come back to this velocity triangle as we proceed to solve this question. So, first part of the question we will what we will try to do is that since we know the pressure drop across the overall pressure drop is known. Let us calculate the overall stagnation temperature drop across the turbine and then what we will do is to estimate the stagnation temperature drop in one stage. Therefore, the overall temperature drop divided by temperature drop in one stage will give us the number of stages and estimate of the number of stages required. So, the overall pressure ratio is given as 6. Therefore, T 0 1 by T naught E s where T naught E s is the stagnation temperature isentropic at the exit of the turbine is equal to P 0 1 by P 0 E raise to gamma minus 1 by gamma. Therefore, the stagnation temperature at the exit of the turbine is equal to 576.9 Kelvin. This is isentropic temperature. So, the actual temperature overall will be equal to the isentropic temperature difference multiplied by the isentropic efficiency which is 85 percent in this case. So, delta T naught overall which is the stagnation temperature across the entire turbine is equal to eta T which is the isentropic efficiency multiplied by T 0 1 minus T naught E s. T 0 1 is given as 900 T naught E s is calculated as 576.9 Kelvin. So, the overall temperature drop in the turbine is 274.6 Kelvin. So, now we come back to the velocity triangles. We know from the inlet velocity triangle C 2 is C a divided by cos alpha 2. C 2 is this cos alpha 2 is C a divided by C 2 and therefore, C 2 is C a by cos alpha 2. C a is known alpha 2 is also known. Therefore, C 2 is equal to 150 divided by cos 75 that is 579.5 meters per second. Now, the stagnation temperature at the inlet is known T 0 2 which is equal to T 0 1 and therefore, static temperature at the nozzle exit T 2 is equal to T 0 2 minus C 2 square by 2 C p. Therefore, T 0 1 which is equal to T 0 2 is 900 Kelvin C 2 we have just now calculated 579.5 and therefore, we can calculate T 2 that is 753.7 Kelvin. Now, it is given specifically that this is an impulse turbine stage which means that the degree of reaction is basically 0. So, when the degree of reaction is 0 it implies that the degree of reaction we have defined as H 2 minus H 3 divided by H 0 1 minus H 0 3. Since, degree of reaction is 0 we have H 2 is equal to H 3 which implies the static temperature should be the same T 2 and T 3 should be same and that is equal to 753.7 Kelvin because there is no change in static conditions across a rotor. The entire pressure drop has actually taken place in the nozzle nothing changes in the rotor T 2 is equal to T 3. Now, at the exit of the velocity triangle at the rotor entry well let us look at the rotor entry first we have calculated alpha 2 which is known to us let us calculate beta 2 as well tan beta 2 we know is C 2 sin alpha 2 minus u by C a. Beta 2 is C 2 sin alpha 2 that is this component minus u divided by C a. So, that is C 2 sin alpha 2 minus u divided by C a and that comes out to be beta 2 comes out to be 64.16 degrees. And therefore, from this we can calculate the relative velocity v 2 and that is C a cos beta 2 that is 344.15 meters per second for constant axial velocity in an impulse turbine we have seen that v 2 is equal to v 3 which is equal to C 3. And therefore, we can calculate this stagnation temperature at the rotor exit T 0 3 in terms of because T 3 and T 2 are same T 0 3 is T 2 plus T 3 square by 2 C p that is 753.7 plus 344.14 square by 2 into C p that is 805.28 Kelvin. Therefore, the per stage temperature drop stagnation temperature drop is T 0 1 minus T 0 3 and that is 900 minus 805.28 94.7 Kelvin. We have already calculated the overall temperature drop which was 274.6 that divided by 94.7 comes out to be 2.89 that is to be approximated to the next integer that is 3 stages. So, we have we are required 3 stages for achieving this kind of a pressure drop in and generating the work output from this kind of an impulse turbine which has this efficiency and mean blade speed and so on. So, this is one of the ways of estimating the number of stages that are required basically calculating the overall temperature drop and calculating the temperature drop per stage dividing the 2 will give us the number of stages required. So, this completes this particular problem as well we have solved. So, for 4 distinct problems which are to do with different types of turbines how to analyze the velocity triangles and solve the problem using these velocity triangles. So, what I will do now is to leave you with exercise problems which you can solve based on what we have solved in today's lecture as well as the discussions we have had in the last few lectures. I will also give you the final answers of these questions. So, you can check after you have solved these problems. So, the first exercise problem is the following an axial flow turbine operating with an overall stagnation pressure of 8 is to 1 has a polytropic efficiency of 0.85. Determine the total to total efficiency of the turbine. If the exhaust Mach number of the turbine is 0.3 determine the total to static efficiency. If in addition the exhaust velocity of the turbine is 160 meters per second determine the inlet total temperature. The answers to these questions are 88 percent that is total to total efficiency total to static efficiency is 86.17 percent inlet total temperature is 1170.6. The second exercise problem is the mean blade radii of the rotor of a mixed flow turbine is 0.3 meters at the inlet and 0.1 meters at the outlet. The rotor rotates at 20000 revolutions per minute and the turbine is required to produce 430 kilowatts. The flow velocity of the nozzle exit is 700 meters per second and the flow direction is at 70 degrees to the meridional plane. Determine the absolute and relative flow angles and the absolute exit velocity if the gas flow rate is 1 kg per second and also the velocity of the through flow is constant through the rotor. So, the angles are alpha 2 is 70 degrees, beta 2 is 7.02 degrees, alpha 3 18.4 degrees and beta 3 is 50.37 degrees. The third question is an axial flow gas turbine stage develops 3.36 megawatts at a mass flow rate of 27.2 kg per second. At the stage entry the stagnation pressure and temperature are 772 kilo Pascal and 727 degree Celsius respectively. The static pressure at exit from the nozzle is 482 kilo Pascal and the corresponding absolute flow direction is 72 degrees to the axial direction. Assuming axial velocity is constant across the stage and the gas enters and leaves the stage without any absolute swirl velocity, determine part a the nozzle exit velocity, part b the blade speed, part c total to static efficiency and the stage reaction. Answers are 488 meters per second, the blade speed is 266.1 meters per second, total to static efficiency 0.83 and the stage reaction is 0.128. And last question is a single stage axial turbine has a mean radius of 30 centimeters and the blade height of 6 centimeters. The gas enters the turbine stage at 1900 kilo Pascal and 1200 Kelvin and the absolute velocity leaving the state stator of the nozzle is 600 meters per second and inclined at 65 degrees to the axial direction. The relative angles at the inlet and outlet of the rotor are 25 degrees and 60 degrees respectively. If the stage efficiency is 0.88 calculate part a the rotor rotational speed part b stage pressure ratio, part c flow coefficient d degree of reaction and part e power delivered by the turbine. Here the answers are rotational speed is 13,550 rpm, the pressure ratio is 2.346, flow coefficient is 0.6, degree of reaction 0.41, power delivered is 34.6 megawatts. So, these are four different problems that I have put up as an exercise for you and I hope with what we have discussed in the last few classes as well as today's tutorial you will be able to solve these problems based on these discussions. So, we will continue our discussion on axial turbines further in basically looking at the 3D flows in further lectures and I hope you will be able to understand the basic working of a three dimensional flow in axial turbine based on the fundamentals that we have discussed based on the two dimensional flow in axial flow turbines. So, we will continue discussion on some of these topics in the coming lectures.