 We are talking about axial flow turbines and in the last class, last lecture, we talked about how to design and bring into the design steps and design features three dimensionality of flow through axial flow turbines. Now, we know that the flow through the axial flow turbines as in axial flow compressors is normally annulus in nature and that flow through the annulus space that is available to the turbines is often not exactly uniform. So, some aspect of the non-uniformity or some aspect of the variation from the lower radius to the outer radius or inner radius to the outer radius and more specifically from the root to the tip of the blade of a rotor needs to be factored into to begin with in the design and later on in the computations and of course, later on in the analysis. We have seen how some of these things can be brought into simple analysis that we had done in the last class without getting into more complex computational analysis. As I have mentioned, we shall do an introduction to computational analysis through turbo machines towards the end of this lecture series, but right now we are looking at turbine specific certain theories which factor in the three dimensionality of the flow and built it into the design. So, most of it is indeed used for design and then immediate post design analysis to find out how the turbine is actually going to behave. So, in today's lecture we will look at some problems that actually use these theories that we have done in the last class and then try to actually solve some problems which are prescribed problems and from the problem statement we try to figure out what kind of solutions can be arrived at using the theories that we have done in the last lecture. And towards the end we I will leave you with a few problems to solve by yourselves. So, that you can get the feel of the application of these theories to actual problems, realistic problems and you also get a feel of the numbers. It is essential for an engineer to get the feel of the numbers and that is why it is essential that we look at a few problems which are probably a little textbook problems simplistic problems where you know typically you have all the data you require in a real life problem you often may not actually have all the data that is required to solve, but we are dealing with problems where the problem statement gives you all the data that you require and now you need to use the theories that we have done to arrive at solutions and in a process one learns how the solutions would look like and indeed as I said you get a feel of the numbers what the numbers would look like. So, that is very important for all practical engineers. So, let us look at some of the problems related to three dimensional flow in axial flow turbines. Now, in axial flow turbines that we are going to do some solve problems and then I leave you with some exercise problems to solve for yourselves. Now, in the solve problems we will first look at the problem statement in the first problem the problem statement shows that a constant nozzle exit angle which we have designated earlier as alpha 2 is being used for axial flow turbine design in which it is prescribed that the temperature drop should be 150 k and at the hub the blade velocity or blade speed u is 300 meters per second and at the tip it is 400 meters per second at it is prescribed that alpha 2 is 60 which is constant as prescribed from a root to the tip hub to the tip and alpha 3 is 0 that is 0 will at the exit of the rotor. The radius ratio given for this particular problem statement is 0.75 that is hub radius to tip radius ratio is 0.75 that is prescribed here. The problem ask for solution in which you should complete the design velocity diagrams at hub mean and tip of the stage and thereafter calculate the velocity components if the design is a free vortex design for the turbine and compare the results of that with that of constant nozzle exit angle. So, one can look at the whole thing from a free vortex point of view and compare the results. So, this is a problem statement which we can now try to find a solution to. Now, at the rotor inlet station we have done in the last class the equation for variation of whirl component is defined by the variation of alpha 2 and that is given by C w 2 by C w m which is the mean and that is also equal to C a 2 by C a 2 mean which is also equal to C 2 by C 2 m which is mean and all these velocity ratios are equal to radius ratio r by r mean r being at any radius to the power sin square alpha 2. Now, this is what we had done in the last lecture following which what we get is at the rotor exit the actual velocity C a 3 square is equal to C a 3 mean square plus twice u m into C w twice m whole thing multiplied by 1 minus radius ratio r by r m to the power cos square alpha 2. Now, this is also the theory that we had the expression we had shown in the last lecture. Now, from the prescribed radius ratio that is given we can see that the mean to tip radius ratio would be 0.875 and mean to hub radius ratio would be 1.166 that is calculated from the hub to tip radius ratio that is prescribed in the problem statement. Now, the work done by the rotor is given by the Euler's equation which you are all aware of. Now, in this particular problem statement alpha 3 would be equal to 0 exit will component is 0 which means C w 3 indeed would be 0 and in which case the specific enthalpy rise or you know specific work that we normally use for aerothermodynamic relations that is C p into delta t would be equal to u m into C w 2 m C w 3 p 0 and from which we can write that C w 2 m would indeed be 492 meters per second and corresponding C a 2 would be from the velocity triangle that you can draw at station 2 that is before the rotor and that would come out to be 284 meters per second as per the prescription that also would be equal to actual velocity at the exit of the rotor C a 3 m. Now, at the rotor hub at the inlet of the rotor one can write actual velocity C a 2 h is equal to C a 2 m to the multiplied by a radius ratio r m by r to the power sin square alpha 2 and this would yield actual velocity of 318.8 meters per second. On the other hand the world component at hub at the inlet C w 2 h would be C w 2 m into r m by r to the power sin square alpha 2 and that would be 552.2 meters per second. So, you now start getting the components of the velocities at station 2 that is before the rotor at the rotor inlet. Now, at the rotor tip of the inlet one can now find the actual velocity using the same relation that is C a 2 m into r m by r to the power sin square alpha 2 and in which case the actual velocity would now be 257 meters per second and the world component C w 2 tip would be 447 meters per second. So, as one can see that the world component has actually decreased from hub to the tip whereas, the actual velocity has also decreased a little from hub to the tip and this is a consequence of the fact that alpha 2 has been held constant from root to the tip of this prescribed problem. So, as a result of which we now have all the velocity components that are required at the inlet to the rotor. Now, at the rotor tip outlet we can find what the actual velocity would be and this can be found from the actual velocity relation that we had done earlier and that is given by C a 3 equal to whole thing root over C a 3 square plus twice u m C w 2 m multiplied by 1 minus radius ratio r by r m to the power cos square alpha 2 and if you do that we can calculate the actual velocities at tip and hub as at the outlet as 262 and 306 meters per second. So, once you apply the relations that we had done in the last class for three dimensional flow estimation we now have the actual velocities at tip and hub at the outlet to the rotor. On the other hand if we calculate all the values with reference to the free vortex design we can see that the velocities that we get in comparison to the nozzle angle can be now written down in a tabular form and one can see that these values are the ones in constant nozzle angle are given in red and the free vortex values are given in black and the actual velocities as you can see in in front of the rotor and behind the rotor are constant for free vortex only the world component is changing as per free vortex variation from tip to hub. On the other hand for constant nozzle angle as one can see all the velocities are varying and essentially only at the mean the values of free vortex and the constant nozzle angle are same at the tip as well as at the at the mean the actual velocity the world component velocity and the exit actual velocity they are all same at the mean but at the hub and the tip the constant nozzle angle gives completely different values compared to that of free vortex. Now, this is something which you may like to take a look at and you may like to sit down and draw all the velocity triangles that are born out of this two calculations two sets of design calculations and you will probably find that you get completely different blade shapes the blade shapes that come out of this two design exercises would indeed be quite different from each other and that would tell you that if you use different kind of design law or design philosophy you would in end up getting quite different blade shapes the three dimensional blade shapes. Now, if we go on to the second example in which the problem statement reads that it is proposed that for design of an axial flow turbine two design methods are to be explored. Now, first design method that is prescribed is C w 2 m is equal to C w 2 h that is equal to C w 2 t that means the world components at hub mean and tip are equal to each other. Now, this type of design is what we have called earlier a solid body essentially design that means the fluid behaves more like a solid body and correspondingly the world components everywhere are same. The second design that is prescribed is where C a 2 is equal to C a 2 hub into the radius ratio hub to tip radius ratio to the power sin square alpha 2 which somehow tries to make use of the constant alpha 2 prescription and that is your case B the second design that is suggested here in case B and in case C we have C w ratio C w 2 t to C w 2 h as equal to the radius ratio r h by r t. Now, that if you remember is nothing but your free vortex design. So, we have three design that is suggested for design of an axial flow turbine one in which the flow behaves like a solid body the second one in which one may use the alpha 2 equal to constant that is constant nozzle angle prescription and case C is which resembles of that of a free vortex design. What is prescribed are some common data and these are the axial velocity at the mean is prescribed as 200 meters per second entry alpha 2 is 60 degrees exit alpha 3 is 0 the degree of reaction is 0.5 at mean and radius ratio prescribed is 0.8 what is asked for is complete the velocity diagrams for all the cases the velocity diagrams are asked for because those are the velocity diagrams based on which final blade shapes would be created. So, the velocity diagrams would give a fairly good idea about the blade shapes that are being created by three cases A B and C. Let us look at how to go about finding a solution to this particular problem statement. From the prescribed data of the example 2 one can calculate that the radius ratio r m by r t would be 0.889 and r t by r m would be 1.11 this is calculated from the up to t radius ratio that is prescribed in the problem and then the world component at mean C w 2 m can be found from the mean velocity diagram that is C A 2 into tan alpha 2 and that gives C w 2 m equal to 346.5 meters per second and it is prescribed that alpha 3 is 0. So, C w 3 m would be 0 now C w 3 m equal to 0 is what we are done in the first problem also where it is prescribed that C w 3 m could be given as 0 or alpha 3 is given as 0. This is a fairly standard prescription for many of the designs because what happens is when the flow is going out of the turbine quite often the prescription often encourages that the flow going out of the turbine does not have any world component because the world component going out of a turbine typically of a single turbine or a multi stage turbine last stage would be quite useless. So, typically of a turbine it is quite often unless you know it is one of the earlier turbines or one of the middle stage turbines the prescription quite often comes with alpha 3 equal to 0 which leaves a 0 world component and if the flow is going into a nozzle or going into exhaust any world component present is quite useless only component that is useful for nozzle effect for thrust making is the actual component. So, giving a prescription of world component 0 is pretty much a practical and standard prescription for turbine design as I mentioned unless you are designing a turbine which is one of the middle stages of a multi stage turbine. So, as we have seen in both the problems C w 3 and alpha 3 have been prescribed to be 0 and as a result of which the problems does become a little simple, but the prescription is realistic it is not really idealistic. Let us get back at the problem for this particular problem statement it is given that degree of reaction R x is 0.5 now that means it brings us back to the symmetrical blading concept that you may have done earlier and certainly done in some detail in axial flow compressors. So, the moment you put a degree of reaction 0.5 the symmetrical blading concept comes in and then you have at the mean alpha 2 m equal to beta 3 m that would be equal to 60 degree as prescribed and alpha 3 m equal to beta 2 m equal to 0 degree as prescribed. This of course, makes the problem a little simple to handle. However, many turbines in the past and the early days of turbine design have been used using these kind of somewhat simpler design prescriptions and those turbines were operating quite well. It makes the designers job definitely simple and probably analysis also becomes simple. In those days long back 30, 40 years back if you may remember the aid from computational flow dynamics was not available and as a result the refinement that is possible in today's turbine design was not really available in those days and as a result somewhat simpler design prescriptions were often used for design and those were functional they worked fine. So, a similar somewhat simpler design prescription has been also prescribed here. Now, the blade velocity u m would come out to be same as C w 2 m because your beta 2 m is 0 and as a result of which it is 346.5 meters per second as calculated just a little above and at any radius we can now calculate the blade velocities from the radius ratios that we have just written down. So, the hub blade velocity would be 308 meters per second and the tip blade velocity u t would be 385 meters per second. So, the blade velocities vary all the way from root to tip as per omega into r concept. Now, for case a we have 3 cases to be actually looked at and as I indicated this is a fluid which is prescribed to be behaving like a solid body for a case which we have done in the last lecture n equal to 0 for the equation C w to equal to r to the power n. Now, when you put n equal to 0 it is a case which fluid behaves like a solid body. Now, the actual speed is calculated from the actual velocity expression derived from the energy equation for the case n equal to 0 and this comes out to be C a 2 is equal to C a 2 m whole thing root over 1 minus 2 tan square alpha 2 m into l n radius ratio r by r m. Now, this can be derived you can sit down on derive it the same way we are done earlier for the case n equal to 0 from the energy equation that we are written down involving the various velocity components. And you have to put the value of n equal to 0 there and you would arrive at this solution which we are looking at for actual velocity at any station with relation to actual velocity at mean radius along the blade length. Now, using this you can also calculate the angles across the rotor from the above considerations. If you do that the solutions that you get for the case a tells you that the actual velocity prescribed at the mean was 200 and C w 2 at the mean was found to be 346.5. On the other hand we found the actual velocities varying from tip to hub and both at the rotor entry as well as at the rotor exit at the rotor exit the C w 3 is prescribed to be actually 0. And correspondingly the value of alpha 3 also has been prescribed to be 0 the alpha 2 variation is shown here as part of our results it varies all the way from hub to the tip correspondingly the beta 3 also varies exactly in the same manner from hub to the tip. And the beta 2 values are shown here it is 0 at the mean as we have calculated as has been prescribed the degree of reaction being 0. However, there is a small value of beta 2 at the hub and a small value of beta 2 at the mean at the tip that has been shown here. Now, if we move to the results of case b the prescribed condition is C a 2 t is equal to C a 2 h into radius ratio r h by r t prescribed here in the problem to the power sin square alpha 2. Now, this of course brings us to the fact that all the velocities at the tip and hub and mean can be the ratios of them can be put down as equal to each other all of them would be equal to some relation to alpha 2. Now, this allows us to write down that for constant nozzle angle which is the prescribed case b that we are looking at C a 2 can be now written down as C a 2 m into radius ratio r m by r to the power sin square alpha 2. And in case of C w 2 that will be C w 2 m to the into radius ratio r m by r to the power sin square alpha 2 and C 2 is equal to C 2 m to the multiplied by radius ratio r m by r to the power sin square alpha 2. So, for constant nozzle angle case all the velocities C a 2 C w 2 and C 2 which is the absolute velocity can be found from the mean values that is at the mean radius to hub to tip at any radius by using this radius ratio concept. If we do that at the station 3 at the exit of the rotor it is prescribed that alpha 3 is equal to 0 and C w 3 is also equal to 0 the expression for actual velocity comes out as we have done in the last lecture is C a 3 square is equal to C a 3 square 3 C a 3 m plus twice u m C w 2 m multiplied by 1 minus radius ratio to the power cos square alpha 2. Now, this allows us to calculate C a 3 at exit of the rotor if we now use the relation that we have done which is essentially for the case B which is exit angle from the rotor is held constant from root to the tip of the blade. Now, this is something which we have discussed in the last lecture that holding the exit angle from the rotor constant from root to the tip of the blade makes the rotor untwisted or very likely twisted. Now, this is important for stator nozzle blade cooling purpose we shall be doing the cooling technology from next lecture onwards, but this particular design philosophy of holding alpha 2 to equal to constant from hub to tip or root to tip of the blade of the stator essentially caters to cooling technology and if we apply this in this present problem what we see is the results that we get for case B. Let us look at the results of case B what we see here is the actual velocity is now varying all the way from hub to the tip at the hub it is 218.5 at the mean it is prescribed as 200. So, the variation of actual velocity at the entry as well as at the exit is quite pronounced from hub to tip the values of C w 2 are also variable from hub to tip quite substantially C w 3 is being held constant alpha 2 by prescription is held constant alpha 3 by prescription is 0. We get a variation of beta 2 from 17.9 to minus 19.4 and we get a variation of beta 3 from 54 to 69 from hub to tip. So, these are the results of the case B which is of a constant alpha 2 from hub to the tip of the stator nozzle. Now we can move to the case C now case C is what we had seen was actually the free vortex design. Now free vortex design as we have mentioned before is not the most popular design for actual flow turbine even though it is a very popular design for actual flow compressors for turbine it is not the most popular design, but it is the simpler design it of course works if you make actual flow turbine with free vortex design it will surely work there is no reason why it should not work, but it is not the most popular design today and ever since the cooling technology came into the market the free vortex design has been essentially replaced by the constant alpha 2 design which is the more popular design philosophy for turbines essentially as I mentioned to cater to cooling technology, but let us look at the results that we get for this problem statement problem 2 case C for free vortex design philosophy. If you apply free vortex design philosophy the world components at tip and hub the ratios are directly related to the hub to tip radius ratio. Now if you apply the same at the outlet also C w 2 also is held constant as per free vortex principle across the blade at mean radius. So, C w C a 2 is equal to C a 3 and if you apply all this in the as per the free vortex law we get a set of results directly as per very well known free vortex law prescriptions the results that we get are that the actual velocities are held constant from hub to tip as well as across the rotor C w 2 varies substantially from hub to the tip as per free vortex law and which would tell you that you would end up getting a substantially twisted blade C w 3 is being held constant. So, the trailing edge would be rather linear on the other hand the value of alpha 3 is 0 corresponding to the prescription alpha 2 varies from hub to tip and the C w 2 variation shows that and then of course, you have the variation of beta 2 which goes minus at the tip and this is something that comes out of the free vortex design and as a result of which you get a variation of beta 3 which also varies from hub to the tip of the blade. So, you have the results of the case C tabulated here and then we have three cases a b and c these variations can now be all put together into one table which essentially tries to compare the three cases a b and c and as you can see the case a is given in red the case b that is a constant nozzle angle is given in b and the free vortex design is given in case c and all three of them are brought together if you sit down use the velocities that are given the angles that are given and if you draw the velocity triangles of all the cases you would probably get a very clear picture of what kind of blades actually come out the three dimensional blade shapes that should come out of the three cases that are prescribed here. The three cases that are prescribed here had certain commonalities that is the mean actual velocity prescribed where 200 the exit angles were prescribed to be 0 the will component of the exit were prescribed to be 0. So, with those common prescriptions we try to put together the blades that would come out even with those commonalities and we see that three completely different blade shapes are likely to result from three cases that are prescribed here for axial flow turbine design. So, as I mentioned you can probably sit down and actually draw the velocity triangles and you would find that three different cases three different blade shapes and you would indeed need to choose different airfoil shapes different blade sections from hub tip for each of these three cases. So, you end up having three completely different blade shapes for three different design philosophies even though we started off with common data prescriptions for all three cases. So, this is an example which tells you with certain simplifications a simplified problem it still tells you that if you have three different design philosophies you end up with three completely different blade shapes. I will now leave you with a few problems which you can sit down and solve for yourself and get a feel of the numbers that come out of solving of examples problems that are prescribed with numerical values. So, let us look at some of the problems you can solve for yourselves. The first exercise problem the problem statement reads that an axial turbine rotor is prescribed with a rotor inlet and outlet flow in radial equilibrium which means the static pressure and the dynamic pressure are balanced at the inlet and outlet of the rotor. The world component of the flow is designed to vary radially as per this prescription as C w at the inlet as a r minus b y r and at the outlet as a r plus b y r. Now, a and b are the constants and what is required for you to find is find the inlet and outlet exit velocities and the expressions for those velocities. In this case you can see the answers given here and the axial velocities of course, would remain constant across the rotor. So, we are dealing with the rotor only. So, the problem statement is essentially for rotor. Part b of the problem is it is prescribed that at mean radius given value is 0.3 meters the actual velocity is 10 meters per second and the degree of reaction is 0.5. The blade loading coefficient is prescribed as per the definition psi rotor is equal to work done specific work done h 0 by u tip square and the rpm is 7640 rpm the radius ratio is given as 0.5 and at 80 percent of the rotor radius it is required for you to find the rotor relative flow inlet and outlet angles. So, what you are required to find the beta values beta 2 and beta 3 for the particular problem statement that is given here. Now, the answers given are 43.2 degrees and 10.4 degrees for beta 2 and beta 3. So, you can try to sit down and see whether you can arrive at those answers. The second exercise problem that is given is the gas exits from the turbine stator or nozzle at a radially constant angle alpha 2. So, it is a constant nozzle exit angle problem. The gas is prescribed to be in radial equilibrium. The actual velocity variation at that station is given as C A 2 into R to the power sin square alpha 2 equal to constant. This is what we have done in our lectures also and for a turbine in which the actual velocity at the radius 0.3 is again prescribed as 100 meters per second and if the turbine as stated above is designed with a constant alpha 2 equal to 45 degree find the actual velocity at that station at 0.6 meters radius. Now, the answer given here is very simple and that is 70.7 meters per second. So, it is a constant exit stator exit angle problem and you have to apply the relations that we have done in the lecture or in the earlier problem that has been solved for you. The third and the last problem that is prescribed for you to solve is an actual turbine is designed with free vortex at stator nozzle exit and 0 whirl at stator at rotor exit. So, it is a free vortex problem applied at stator nozzle exit and a rotor inlet. So, the station between the stator and the rotor carries free vortex prescription. For the following operating condition that is at the entry T 0 1 is equal to 1000 k, mass flow is given as 32 kg per second, the hub radius is 0.56, the tip radius is 0.76 meters, rpm prescribed is 8000, the degree of reaction is 0.5 and actual velocity is constant that is 183 meters per second and it is prescribed that the inlet and exit absolute velocities are equal to each other that is C 1 is equal to C 3. You are required to find C 2 that is the nozzle exit velocity as you know it expands hugely from C 1 to C 2 and then you are required to find the maximum Mach number at the stage. In this particular stage typically it is most likely to be somewhere in the nozzle exit and then the reaction at the root, the power output of this particular working turbine and T 0 3 and T 3 at the stage exit. So, these are the numerical values that you need to find out of this prescribed problem. The answers solutions that are given here is that the nozzle exit velocity C 2 is would be 480 meters per second, the maximum Mach number in this stage is solved as 0.818, the reaction at the root is 0.08 and you remember at the root it is quite often especially if it is free vortex it is likely to be very close to 0 and 0 of course would mean an impulse turbine and we are looking at a problem in which the solution actually comes pretty close to giving you an impulse station or impulse section at the root of this particular turbine and then the power output or work done for this particular rotor given the mass flow is 3.42 megawatts and the temperatures at the exit T 0 3 is 907 and the static temperature T 3 is 892 k. Now, you can sit down and try to solve this problem and see whether you can come. You can use standard values of gas constant R that is 247 joules per kg k and the value of C p as 1147 joules per kg k. So, you can use those standard values to solve this problem in which numericals are given and you are required to find the certain prescribed velocities, Mach numbers, work done and the exit temperature. So, I will leave you with these problems for you to solve for yourself. So, that you can get a feel of the numbers that typically come out of actual flow turbine design. So, some of these problems would give you an idea how the turbine design is indeed proceeded with and what kind of numbers you get, what kind of variations you get, you get a feel of the numbers by solving these problems. In the next class, we will be looking at turbine blade cooling because in this class and in the earlier lecture, we had looked at the design philosophy of alpha 2 is constant from root to the tip and we have stated again and again that this particular design philosophy essentially caters to turbine blade cooling. In the next lecture onwards, we will devote ourselves to looking at this turbine blade cooling technology and how it impacts the turbine blade, the turbine blade shape and essentially the aerodynamics or aerothermodynamics of the flow over the turbine blades is very strongly impacted and essentially the aerodynamics of the blade changes hugely by application of cooling technology. We will look at various cooling technologies and how do they actually impact the turbine design of modern actual flow turbines. Specifically, these cooling technologies are very widely used in aero engines and we will look at some of the typical examples of these applied cooling technologies in turbine, actual turbine, rotor and stator. So, we shall be doing turbine cooling technologies from next lecture onwards.