 We are talking about gas turbines and we just had a look at the basic theories of axial flow turbines and radial flow turbines. In today's class, we will take a look at how simple problems related to these two kinds of turbines can be solved using the theories that we have done. So, I will probably solve a couple of problems for you and then I will leave you with some problems for you to solve for yourselves using the gas turbine theory that we have done in the last two or three lectures. Now, these problems are set out for you, so that one, you can use the theory that we have done in the last lectures and secondly, you get a feel of the numbers. In engineering, it is extremely important that you not only understand the equations and the theories, that you also get a feel of the numbers. What are the possible correct numbers under various situations for various parameters, that feel is extremely important for good engineers. So, hopefully in today's class, I will be able to expose to you some of these engineering quantitative issues and you get a feel for the numbers. So, in today's class, we will do the tutorials, solved examples and tutorial problems on axial and radial flow turbines. Now, the radial and axial flow turbine that we have done, we know are slightly different kinds of turbines even though they do the same kind of job basically that they perform work. So, first we will take on the axial flow turbines and take a look at a problem which represents a typical axial flow turbine problem and then we will discuss some of the tutorial problems that you could possibly attack for yourselves and get a feel for the numbers. The first problem that we have is on axial flow turbine and this problem statement shows that if you have a cooled axial flow turbine. Now, we have discussed the cooling technology of axial flow turbines. So, if we and most of the modern axial flow turbines do tend to be cooled turbines unless it happens to be the last stage of a multi-stage turbine. So, let us take a look at a typical cooled axial flow turbine problem and the problems shows that it has properties such that its entry temperature is 1780 k and entry pressure is 1.4 megapascals and carries a mass flow of 40 kgs per second which means it is a medium sized engine. The mean radius data that means the mean is between the hub and the tip of a blade. So, the mean there is given as Mach number the entry absolute Mach number is given as 0.3. The exit from the stator is given as 1.15 which means the flow has indeed gone a little supersonic at the exit of the stator and the rotational speed at this mean radius is 400 meters per second. T03 that is the exit temperature from the turbine is given as 1550 k and some simplistic angular parameters are prescribed here that is alpha 1 is equal to alpha 3 is equal to 0 degree that means flow is coming in and going out axially that of course simplifies the problem, but it is also reasonably realistic. So, it is not just ideally simplified, but it does represent a realistic flow situation and then at the mean the radius is 0.4 meters and as I mentioned it represents probably a medium sized engine. The axial flow velocity ratio across the rotor Ca 2 by Ca 3 is equal to 1 that means the actual velocity across the rotor is conserved. This is again a simplifying assumption, but also a reasonably realistic assumption it is not very ideal it is a realistic assumption. The loss coefficients through the rotor and the stator the stator is of course, we have been calling it a nozzle. So, the loss coefficient across the nozzle is given as 0.04 and that across the rotor is given as 0.08. So, these two are prescriptions these numbers are realistic numbers and represent a reasonably real compressor turbine situation. Given other thermodynamic parameters the specific heat ratio gamma is given as 1.3 and the gas constant R is given as 287 J per kg k. Now, those are standard parameters and they have been prescribed for this problem also. Now, the problem asked for you to compute the flow properties along the mean line of the stage the degree of reaction the total temperature delta T 0 based on stage loading a based stage loading. Now, this is something which we have discussed before. So, we will try to compute this and then the isentropic efficiency which has been defined and the flow areas at various axial stations and at the hub and tip radii at the nozzle inlet at the nozzle exit and at the rotor exit. So, let us take a look at these problems the parameters that are given and see what kind of solutions we can get out of this problem that has been stated here. Now, we try to put the problem statement in terms of at the mean radius as it has been prescribed in terms of the velocity diagram that we have done before and the problem statement now cast on to the velocity diagram that we are familiar with shows that the entry temperature at station 1 has been prescribed 1780 k and pressure has been prescribed. Now, at the station 1 alpha 1 there is now prescribed as 0 degree and the entry mark number 0 3 corresponds to C 1 which is shown in the diagram. So, that C 1 corresponds to M 1 equal to 0.3 at that particular station. Now, at the exit from the stator nozzle the mark number prescribed is 1.15 which is corresponding to the velocity C 2 that is the absolute velocity and that absolute velocity is prescribed in the form of mark number as 1.15 depending on the local flow condition that is local temperature and the corresponding rotating speed u is given as 400 meters per second. So, in that velocity diagram you can see the value of u and that number quantitative number is given as 400 meters per second. The other thing that is prescribed is across the rotor the actual velocity ratio. Now, you can see C A 2 and C A 3 in the velocity diagram here and that ratio is prescribed as 1 in this problem statement. Also prescribed is that the exit absolute flow angle alpha 3 is 0. Now, this is something which is similar to the entry absolute flow angle. Now, corresponding to the prescription that has been given here we have the loss coefficients the loss coefficients prescribed for the nozzle or the stator nozzle blade is 0.04. The similar loss coefficient that is prescribed for the rotor is 0.08 normally you would see that the rotor losses are somewhat on the higher side because it is a rotating blade and in rotating blade the losses are normally of various origin and this is something we have discussed a little and you would probably appreciate the fact that the number related to the rotor is on the higher side actually it is almost double of that of the loss in the nozzle. The other thing that is given is that the exit total temperature is 1550 k which means it has undergone a total temperature drop of the order of 230 degree k across this turbine. That is the amount of work you can expect that this turbine would have probably accomplished in the process of functioning or in the process of transfer of energy from the gas to the rotor and then mechanical work. Now, this is something which is what the turbine exists for and the quantitative value prescribed here is 230 k that is a pretty large number and you would probably quickly notice that that number is substantially higher than what you can get in terms of temperature rise in any compressor stage. So, these are the prescribed values for the problem that has been put in front of us. So, let us go along and see whether we can complete the process of solution of this particular problem. Now, at station 1 if we look at the numbers that are given and try to find the static parameters. If we can get a value of T 1 and if we apply the isotropic flow conditions at station 1, we get T 1 is equal to 1756.3 k using the isentropic flow conditions and P 1 correspondingly using the same isentropic conversion from total to static condition. We get 1321 kilopascals converting from megapascals to kilopascals. Now, we are using isentropic flow condition because as we have done before in various thermodynamics that at any particular station conversion from static to total or total to static can be simply by using the ideal isentropic relations because at that particular station it is assumed that the flow is undergoing an isentropic conversion from total to static or static to total. That is an assumption that is of course, the theory and we can always invoke that theory at a particular station. There is no process from 1 to 2 it is only at a particular station 1. So, we can always invoke the isentropic flow condition at that particular station in an ideal manner. So, that is what we have done and we have got the static temperature and static pressure at station 1. Now, we can find the velocity C 1 as I mentioned M 1 that was given is corresponding to C 1 and that is dependent on the local temperature that we have just found and if we invoke the local temperature that has been just found we get a velocity which is 242.8 meters per second corresponding to 0.3 that has been prescribed there. Now, since alpha 1 is given as 0 C 1 effectively becomes equal to C a 1 and correspondingly C w 1 would be 0. So, if you keep your eye on the velocity triangle you would see that we are trying to solve all the parameters that are shown in the velocity triangle as far as possible and the trick of solving such problems is that if you can find all the parameters that are shown in this velocity triangle you would be able to calculate the performance of the turbine completely and fully. So, one of the things that you may like to do while solving a problem is to compute all the parameters that are shown in this velocity triangle. So, let us do that and we found that station 1 C w 1 would be equal to 0. Correspondingly then at station 1 we can say that the area a 1 would be equal to the mass flow M dot gas divided by the local density rho 1 into C a 1. Now, this of course, means that we are simply using the continuity condition at station 1 and simply using the continuity condition we can find out the area at station 1. You would notice that you would have to calculate the density rho 1 corresponding to T 1 and P 1 that we have just calculated. So, if you calculate rho 1 and plug in that value here you would get an area that is 0.0628 meter square at station 1 that is at entry to the turbine stator nozzle blade. Now, applying the same isentropic method of conversion from total to static we can find the temperature at station 2 because through the stator nozzle a lot of acceleration has taken place and the total temperature has remained constant, but static temperature has undergone a large amount of change and that shows up here that the static temperature now is 1485.4 k substantially lower than 1756.3 k that we have found at station 1. So, that is a drop due to the large change from potential to kinetic energy through the stator nozzle and correspondingly we find a C 2 using the same method again that is conversion from Mach number to velocity using the local temperature which we have found as 1485.5 and we get a velocity now corresponding to Mach 1.15 velocity of 856.1 meters per second. Now, this is a very high velocity jet and intended purpose of course, is that this would now be made to impinge on the rotor blade to make the rotor rotate. Now, as prescribed 4 percent of the kinetic head is lost in the nozzle blade the nozzle loss coefficient was prescribed as 04 and that is corresponding to 4 percent of the kinetic head as per the definition of the loss coefficient and as a result of which the we can sort of do a reverse engineering and back calculate and find that the ideal T 2 would have been T 2 prime would have been 1503.5 k and corresponding to that the ideal C 2 prime would have been 877 meters per second. Now, this is something we have discussed before in our lecture that through the nozzle because of the real flow situation the actual velocity C 2 is often a little less than what could have been the ideal velocity C 2 prime. Now, this change from real to ideal is quite often a situation that happens because you see at the end of the nozzle you are trying to achieve a sonic velocity sometimes as we see in this particular problem you are trying to achieve a slightly supersonic velocity. So, what happens is you end up getting a little less than what you prescribed or what you design for and hence the loss coefficient comes into the picture over here and this slight loss of a performance through the stator nozzle needs to be accurately factored in your calculation. So, that you know exactly how much you are losing in terms of the acceleration that the stator nozzle is expected to achieve. So, that is what we see here that the ideal velocity could have been a little more through the stator nozzle and as it happens we are getting a little less of the order of almost 20 meters per second, 21 meters per second. Now, this is something what every turbine designer needs to keep an eye on. Now, from the isentropic loss we can also find the value of static pressure P 2 at the end of stator nozzle and since P 0 2 is kind of only factors in the 4 percent loss it tells us that P 2 would be 625.5 kilo Pascal's. Now, at station 3 it is given that exit angle is 0, alpha 3 is 0 and at station 3 it is prescribed that alpha 2 would be corresponding to sin inverse psi into u divided by C 2. Now, this is from the definition of the work done which we have done before that is psi is equal to delta of enthalpy change across the stage divided by u square which is the rotating speed. So, it is used as a normalizing parameter and this is the definition of blade loading or the stage loading quite often used in case of turbines and also sometimes in case of compressors. Now, if you do that you get a parameter that is C a by u into tan alpha 2 from which we wrote down as above that you can get a parameter in terms of alpha 2. Now, C p of course from thermodynamics we know as gamma r divided by gamma minus 1 and if we plug in all those values the stage loading psi can be computed as 1.7878 and this is of course often as a non-dimensional parameter and this number is often an indication of the amount of loading that you can prescribe for a particular stage higher this number is higher the stages being asked to perform and. So, keeping an eye on this number tells you whether the stage has been sufficiently loaded or the stage loading has been maximized and in case of aircraft gas turbine engines as we have discussed before we would like to maximize this number without causing undue gas dynamic or structural problems. Now, from all these calculations what we get is alpha 2 is equal to 56.6 degrees this is the angle at which the flow is coming out from the stator nozzle correspondingly the C a 2 the actual component of this velocity would be 470 meters per second and C w 2 that is the world component or tangential component of the absolute velocity that would be 715 meters per second and of course your v w 2 that is the relative tangential component would be 315 meters per second. So, some of these are numbers which you would probably like to get used to because as you see these numbers are substantially different from what you found in station 1 where C w 1 was of course 0 and hence the change across the stator nozzle only even when no work has been done a substantial change in the velocity field has been achieved through a large acceleration and this is what was of course intended by the so called stator nozzle. If we move forward with these numbers and move on to the rotors what we find there is that at the entry to the rotor the relative flow angle beta 2 can be now found from the tangential velocity v w 2 and the actual velocity C a 2 and if we plug in those numbers we get beta 2 is equal to 33.8 degrees. Now, it can be also shown that corresponding to this that the m 2 relative at the entry to the rotor that is relative Mach number to the entry to the rotor would be 0.76 which is subsonic. Now, we just saw that the exit from the stator m 2 is actually supersonic that was prescribed as 1.15 and we see by conversion through velocity triangle the m 2 relative is now subsonic. Let us take a quick look at the velocity triangle you see corresponding to our calculations C 2 has been prescribed clearly as supersonic we got the velocity value and we got the v 2 velocity value when we converted v 2 to m 2 relative this is clearly subsonic by the local flow conditions and this is exactly what has been achieved through this velocity vector transformation that the exit velocity here can be clearly supersonic, but the entry here would be subsonic and as I have mentioned earlier this is done deliberately that if most of the modern turbines today still admit flow into the rotor which is subsonic. It is indeed possible to admit flow into the rotor which is supersonic that means you can indeed have supersonic rotors, but normally in aircraft gas turbine even today it is not a done thing supersonic rotors have been used in other applications notably in termination the applications in space crafts, but they are normally not used in aircraft gas turbines. So, in aircraft gas turbines we still keep the v 2 value subsonic this is deliberate and very done very consciously for the turbine design. So, that the losses through these turbines are somewhat under control as you can see the loss to the rotor is already little on the higher side and through the shocks it would have gone even higher and hence most of the aircraft gas turbines even today are essentially subsonic rotors. Let us get back to the calculations that we were doing through the rotor we can now find the area at the station 2 which is at the entry to the rotor and that is again found by using the continuity condition and the continuity condition tells us that value of a 2 can be written down in terms of the various flow parameters there and it comes out to be of the order of 0.58 meter square. Now, axial velocity at the exit of the stage can be found simply by using the velocity triangle that we were looking at and this comes out to be 470 meters per second. Now, we have seen that this actual velocity is constant across the rotor. So, the value of C a 2 would be exactly same. So, that is how this velocity has been computed. Now, at the exit again it is been prescribed that alpha 3 is 0. So, C w 3 would again be 0 and therefore, the value of V w 3 would be equal to the rotating speed of the rotor and that is 400 meters per second and corresponding relative velocity leaving the rotor would be 617 meters per second. Now, this is of course, relative velocity. So, as we can see that the relative velocity here is also essentially subsonic and we will see that it is a little on the higher side compared to V 2. Now, the exit flow angle beta 3 can be computed and this comes out to be 40.3. This is invoking the fact that alpha 3 is indeed 0 and static temperature at station 3 now can be computed again invoking the isentropic laws as we have done in station 1 and 2 and if we do that we find the temperature to be 1461 K. The total temperature of course, was prescribed. So, it is easy to use the isentropic laws and find the static temperature that is 1461 K and the sonic speed at that particular station is 738.3 meters per second and we can very well see that this velocity V 3 would be subsonic. So, the exit velocity from the rotor also is subsonic. So, both the entry and the exit velocity from the rotor are deliberately kept subsonic to avoid shock losses which again would reduce the efficiency of the turbine, increase the losses and reduce the efficiency and most of the designers present day designers are extremely sensitive to efficiency of the turbines and hence they try to keep the turbines subsonic even today. And the other thing is most of the work done is being accomplished higher level of work done is being accomplished by increasing the turbine entry temperature and this increase of turbine entry temperature is quite often the focus of attention of modern research and more and more cooling technologies applied and we have prescribed that this particular problem actually applies blade cooling and that is how that high entry temperature was admitted and hence we see that we have a cool turbine here which is a cool turbine the entry temperature is very high and that allows the designer to keep the rotors subsonic and hence the rotors are not under any compulsion to go supersonic to get more work done rotors can be kept subsonic. Another reason why modern designers would like to keep the rotors subsonic is because many of the modern rotors are also subjected to cooling technology. Some of these cooling technologies as we have seen are extremely intricate and if we have supersonic flow over them it will be an extremely complicated fluid mechanic situation and most of the turbine designers would like to avoid that kind of complication in their turbine design because it would impact on the efficiency or aerothermodynamic efficiency or what we call isentropic efficiency of the turbine quite adversely. So, as we see by the numbers that they are deliberately kept subsonic. Let us proceed with this problem and see whether we can complete it. The degree of reaction that we have prescribed before as we see now if it is computed using all the parameters comes out to be only 0.106 and as we have discussed in many of the modern turbines the degree of reaction values are of that order much lower compared to what you have done probably in your compressor chapter where the degree of reaction values are somewhat much higher than these numbers. So, you got to get the feel of these numbers that degree of reaction in turbine is substantially lower than that of compressors and these are the numbers you would need to get used to. The Mach number at the exit then the absolute Mach number is 0.6375 and the relative exit Mach number as we just saw is subsonic and it is 0.836. It is slightly higher than what it came in with into the rotor. So, the rotor is clearly a reaction turbine it has accelerated the flow a little and has produced a reaction and hence it is a reaction turbine. The relative total temperature and pressure at the exit can also be computed to complete the calculation and we get T03 relative as 1614 K and P02 relative as 897.3 kilopascals by applying the rotor loss. I need to mention here very clearly here that the rotor loss coefficient that was prescribed should be applied across the rotor in a relative frame. So, the 8 percent loss that was prescribed should be applied to P02 relative and if you do that you get P03 relative at as 872.7 kilopascals. Now, this is something you need to again very clearly understand that the loss prescription that is normally given should be applied in the absolute scale in the stator nozzle and should be applied in the relative frame in the rotor and when you are calculating the pressure losses respectively across these two rows of blades. So, losses in the rotor take place in the relative frame in the stator it takes place in the absolute frame. So, you need to take note of this and do your future problems accordingly. Using the isentropic relation we can now find the stator static values that is P3 as 566 kilopascals and the total pressure is 731 kilopascals. Correspondingly we can find the exit area at station 3 and this exit area is now a 3 equal to again using the continuity condition and using m dot which is constant across the stage the local density rho 3 and C a 3 the local actual velocity C a 3. If we apply all those we get at station 3 the area is equal to 0.63 meters per second you can see that the area across the stage is going slightly higher because the flow is actually losing its pressure and temperature and it is losing its density and hence the area requirement would be higher as you go through the stage. So, these are the numbers you need to understand and made I am trying to make it clear that when you get these numbers when you are solving a problem certain numbers would tell you whether your solution is proceeding along the right path and you get the feel of the number that you are probably when you get the numbers in hand. Let us finally, put them all together the performance parameters that we get from this in terms of temperature ratio the total temperature ratio is 1.148 the total pressure ratio or the pressure drop across the stage is 1.915 which is a modest total pressure ratio across turbine stage many of the modern turbines could possibly have much higher pressure ratio across its stage. The efficiency of this particular turbine can now be computed using the thermodynamic laws that we have done before and it simply is 92.9 percent which is a fairly good efficiency many of the modern turbines can have efficiency slightly higher than these, but 92.9 nearly 93 percent is definitely a good efficiency. At each of these stations the blade height can also be calculated simply using the geometry at that particular station that is H i equal to A i divided by twice pi or mean at that particular station and if we put them all in table in station 1, 2 and 3 we get the areas which we have computed before and these areas are now shown over here and then across the rotor as you can see there is been an increase in area and across the stage indeed there has been a slight increase in area. The height as you can see again has changed across the stations correspondingly the tip radius and the hub radius can be found using the geometry that is available with us. So, this table gives us an idea about the changes that typically occur across a stator and the rotor this may vary from one kind of turbine to another if you go from HP to LP this variations would be slightly different. If you go across to the last stage of a LP it will be again slightly different it depends on the turbine design and as I just mentioned that the pressure ratio 1.95 915 is a modest pressure ratio many of the modern turbines could indeed have pressure ratios are quite higher than this and if they are then all the values that you see here in terms of area height and the tip radius and so on would indeed change from station 1 to 2 to 3. So, it depends on turbine and every turbine would have unique such numbers except that if you get the feel of the numbers you would know whether the numbers are coming out in a correct trend and that trend is what you can get the feel of when you are solving a problem, but the numbers for every turbine would be quite different from each other. What I will do now is I will prescribe a few problems for you to solve for yourself using the methodology that we have just solved the problem. So, the first problem is actually that of a impulse turbine where reaction is 0 and if we do that we prescribe that it has a entry pressure of 414 kilopascals static pressure of 207 and exit pressure of P 0 from the stator nozzle of 400 kilopascals. So, it is lost about 14 kilopascals across the stator nozzle the exit static pressure from the whole stage is given as 200 kilopascals. Now, when operating with a rotating speed you mean at the mean of the rotor blade 291 meters per second the entry temperature given is modest 1100 k and alpha 2 that is exit from the stator nozzle is prescribed as 70 degrees that is from the exit of the stator nozzle. Again it is assumed or prescribed that C 1 that is entry velocity is equal to the C 3 which is exit velocity from the stage what is asked for is compute the total efficiency of this particular stage prescribed are the values of C p and gamma which are the normal values the C p is 1148 kg per kilo joules kg per k and gamma is 1.333. So, if you use those numbers you would probably get an answer close to 91 percent as efficiency of the stage I hope you can sit down and solve the problem using the method that we have adopted before. Let us go on to the next problem which corresponds actually now to not a impulse turbine. Let us look at the problem and we will see that the actual velocity as prescribed through the actual turbine is held constant by design that means C a 1 is equal to C a 2 is equal to C a 3 and this entry and exit velocities are also actual that means C a 1 is equal to C 1 C a 3 is equal to C 3 it is a simplified problem and the flow coefficient that is C a by u through the rotor is 0.6. Now, if you do that the gas leaves the nozzle with alpha 2 equal to 68.2 that is at the exit of the stator nozzle. Now, at all mean diameter that means across the stage if you if you just traverse through the mean diameter from the stator entry to the rotor exit you are required to find the stage loading coefficient psi the relative flow angles at rotor mean diameter beta 2 and beta 3 those are the relative flow angles the degree of reaction r x and the total to total and total to state static efficiencies that we have prescribed in our lecture earlier and you are required to find both the efficiencies and the answers given are that your stage loading coefficient is 1.5 the relative flow angles are 40 and 59 degrees the degree of reaction is 0.25 and the two efficiencies the total efficiency is 90.5 and the total to static efficiency is 81.6 as we have discussed during the lecture the total to static efficiency is normally lower than the total to total to total efficiency. So, these are the answers you would get by solving the second problem. Let us look at the third problem this is following the design data applied to an uncooled axial flow turbine you get P01 equal to 400 kilopascals T01 equal to 859 k and at mean radius alpha 2 equal to 63.8 degree degree of reaction is 0.5 and phi the flow coefficient as 0.6 P1 static temperature at the exit at the entry is 200 kilopascals and total to static efficiency is now prescribed in this problem is 85 percent. If the actual velocity is held constant as in the last problem through the whole stage you are required to compute the specific work done by the gas the blade speed and the stage exit static temperature. Now, these are the numbers that you would get you would get 131 kilo joules per kg and the blade speed would be 301 meters per second and the stage exit static temperature would be 707.5 k. If you look at the fourth problem it is an axial flow turbine now again with a cooled rotor now as I mentioned you could have cooled rotor. So, you have a problem here which says that the mass flow is 20 kgs per second the entry temperature total temperature is 1000 k and P01 is 4 bar C a is 260 meters per second and this is assumed to be constant through the entire stage again it is prescribed that C1 is equal to C3 you mean is 360 meters per second alpha 2 that is exit from the stator nozzle is 65 degree and alpha 3 at the exit to the stage is 10 degree and nozzle loss coefficient is prescribed as 0.05. You are required to compute the relative flow angles at the rotor mean diameter beta 2 and beta 3 at the mean the degree of reaction r x the stage loading coefficient psi the power output of this working turbine and the nozzle exit throat area neglecting the real flow effect that means you assume that it is an isentropic ideal flow and you can still find the nozzle exit throat area and if you adopt all these prescriptions the answers that you would be getting as the relative flow angles would be 37.2 degrees and 57.4 degrees respectively the degree of reaction would be 0.29 the stage loading coefficient would be a very good 3.35 power output correspondingly would be 43.40 watts and the nozzle exit throat area would be 0.04 meters per square. This is the throat area at the exit of the nozzle which is typically at the end of the converging channel that is prescribed or normally used in the stator nozzle. The fifth problem that is prescribed to you is the axial flow turbine with cooled nozzle and rotor blades that means both nozzle and rotor are prescribed to be cooled operates with the following flow parameters at the reference diameter which is the mean. The entry temperature is a respectable good modern temperature that is 1800 k the entry pressure is 1000 kilopascals C A 3 by C A 2 as we have done in our problem is equal to 1 that is axial velocity across the rotor is conserved mark number at the exit of the stator nozzle is 1.1 that is it is allowed to go slightly supersonic and u mean is 360 meters per second alpha 2 prescribed is 45 degree and alpha 3 at the exit of the rotor is prescribed as 5 degree. You are required to compute the following that is C 2 at the exit of the stator nozzle C A 2 the actual velocity between the stator and the rotor C w 2 the absolute tangential component of the velocity at station 2 between the stator and the rotor C 3 the absolute velocity at the exit of the rotor C A 3 the actual velocity and C w 3 the tangential velocity at the exit of the rotor. So, all these velocities that is you are required to complete the velocity triangle that we have seen before the total temperature drop across the stage and correspondingly of course, the total temperature ratio across the stage the pressure ratio across the stage and then P 0 3 for a polytropic efficiency of 89 percent. And if you adopt all the prescriptions that are given and the velocities that are and other parameter that are asked for you get those numbers the velocities are 830 meters per second 585 meters per second and 587 meters per second for C 3 and 585 meters per second for C A 3 and C w 3. You get a temperature drop of the order of 184 K corresponding temperature ratio is 1.148 and the corresponding pressure ratio or a pressure drop across the turbine is 1.792 which gives you exit pressure of 591 kilopascals. So, those are the answers you are likely to get if you adopt the prescription that are given in this problem. Let us now take a look at a problem of radial flow turbine. This radial flow turbine we which we have done in the last class using the simple theory that we have done in the last class we can try to solve this problem. What is prescribed here is that the mass flow is 2 kg per second which means it is a small turbine. The entry pressure is 400 kilopascals the entry temperature is 1100 K and the entry pressure undergoes a 1 percent loss that means the pressure at the exit of the stator nozzle is 0.99 times P 0 1. The nozzle exit angle alpha 2 is 70 degrees. The polytropic efficiency is prescribed as 0.85 for the turbine. The rotor maximum diameter that is at the tip of the radial turbine is 0.4 meters. The radial flow entry into the rotor is prescribed and that is assumed to be or allowed to be assumed to be equal to the exit of the rotor which is C A through and it is supposed to be axial. So, the axial velocity at the exit is to be taken as equal to the radial velocity relative radial velocity at the entry to the rotor. The up to tip radius ratio is 0.4 and T 0 3 at the exit is 935 K. The thermodynamic parameter is prescribed here gamma equal to 1.33 or equal to 287.287 kilo joules per kg K and C P as 1.158 kilo joules per kg K. The problem asked you to compute a rotor tip speed rotational speed and rpm of the rotor. The Mach number velocities the rotor width at the tip of the rotor and T 0 2 relative. The stagnation pressure Mach number hub and tip radii at rotor exit at the rotor exit plane the relative velocity v 3 T 0 3 relative relative flow angle beta 3 and the relative Mach number m 3 at mean radius of course, they are all connected to each other. The values of beta 3 the relative flow angle and m 3 relative at different radii at rotor exit. Now, these are the answers that are expected out of this problem statement. Let us look at the radial flow turbine diagram that we have done before what is prescribed is the tip diameter of the rotor is prescribed as 0.4 at the entry to the turbine what is prescribed is 0.0 P 0 1 equal to 400 kilo pascals T 0 1 equal to 1100 K and alpha 2 over here is prescribed as 70 degree v 2 R is that in this v 2 R here is to be equal to this C A 3. So, this radial component which is equal to v 2 here in this diagram as is normally often done in many gas turbines would be taken as equal to C A 3 that is this C A 3 that is coming out of the rotor the absolute component of the velocity and its axial direction. The T 0 3 prescribed is 935 K and the hub to tip radius ratio of the rotor at the exit station you see there is a hub here station and there is a tip of the rotor at the exit and this is prescribed as 0.4. So, these are the prescriptions with which we go forward the loss across the stator nozzle is 1 percent. So, with this we move forward to solve the whole problem the rotor tip speed can be simply found by using the isentropic laws and that tells us that if you use the enthalpy drop you get a rotor tip speed of 437 meters per second. Now, this rotational speed then can be found simply by dividing this u by r and that gives us to 285 radians per second corresponding to which you would get the rpm equal to 20870 rpm. So, as you can see here a typical radial turbine would be running at very high rpm compared to an axial flow turbine. The corresponding at the rotor tip the value of C 2 would be 465 meters per second using the velocity triangles that we have just seen the corresponding V 2 or would be 159 meters per second. The local speed of sound using the local temperature T 2 you have to do that and if you do that you get T 2 equal to 707 K and corresponding sonic speed as 620 meters per second. And then the nozzle exit Mach number M 2 comes out to be 0.75. So, as you see here it is not sonic it is still subsonic. The area at the rotor tip now is can be found using the continuity condition as we have always done before and this comes out to be 0.0164 you have to find the local density and before that the local pressure. So, if you do that you get this area at station 2 using this area at station 2 which is the entry to the rotor you can find the width of the rotor tip that is the actual width of the rotor tip using the geometry you can find the rotor tip width to be 0.013 meters. This is simple geometry and then you can find the relative total temperature using the flow condition that we have used before that is T 0 2 you take out the absolute flow kinetic energy and plug in the relative kinetic energy and if you do that the T 0 2 relative is 0.0171,017 k. The expansion ratio of the turbine at this operating point using the prescribed polytrophic efficiency can be found by using the relation that we have done before and it gives you a velocity value of 2.16 which of course gives us a P03 of 185 kilo Pascal's. Now given that V2R is equal to CA3 which would be then equal to 159 meters per second and this of course is same as V3 that we can therefore get M3 relative that is the relative Mach number at the exit of the rotor as 0.267 you would of course need to find T3 over there. So, which is taken to be constant from root to tip at the rotor exit. Now at station 3, A3 can be found as using the continuity condition as m dot divided by rho 3 into CA3 and this yields rotor exit area to be 0.02363 meters square. So, that is the area you would get at the exit of the radial turbine rotor. At the rotor exit area if you find the radius using the area that we have found we get the tip area using 0.4 as the ratio tip radius to be 0.0946 meters and hub radius to be 0.378 meters and hence the mean radius would be 0.606624 meters corresponding to which they are CW3 the tangential velocity which is equal to U3 would be 144.8 meters per second. The corresponding velocity triangle at the rotor exit would yield that C3 at the mean of the rotor exit between that is mean between the hub and the tip of the rotor exit is 215 meters per second and the T03 at that mean would be 944 K. The corresponding mean Mach number would can be now computed and you would get a value of 0.362 as the mean Mach number at the exit. The corresponding exit flow angle would be 42.3. The radial variation of these two parameters that is beta 3 and M3 relative can now be found and a Mach number and the exit flow angles go up with radius in a marginally non-linear manner. So, let us look at the solution in a graphical manner beta 3 is shown over here and M3 relative on the other y-axis and the change of radius across as we can see M3 varies along this line path and beta 3 varies in a slightly different manner both of them of course going up from hub to the tip of the exit of the turbine. So, this is what the solution of a radial flow turbine is I will leave you with a tutorial problem which you can solve on your own. The radial turbine problem prescribed here is that the operating point data are given to you P01 is 699 kilopascals, T01 equal to 1145 K. The prescribed nozzle exit pressure is 527.2 kilopascals and the nozzle exit temperature is 1029 K. The stage exit static pressure is given as 384.7 kilopascals and static temperature there is 914.5 K. The P03 value correspondingly is prescribed as 924.7 K. If it is given that the impeller exit area mean diameter to the impeller tip diameter is chosen as 0.49 that is the ratio between tip and mean the design rpm is prescribed as 24000 rpm assuming that the relative velocity at the rotor inlet is radial as we have done before and the absolute flow at the rotor exit is axial again as we have done before. You are asked to compute the total to static efficiency of the radial turbine, the impeller rotor tip diameter and the loss coefficients in the nozzle and the rotor using the definitions of the loss coefficient that we have done in the lecture. If you do those things you would get answers the total to static efficiency would be 90.5, the impeller rotor tip diameter would be 0.27 meters and the loss coefficients would be for the nozzle stator nozzle 0.05, 316 and for the rotor 0.2. As we see the rotor loss coefficient is substantially higher than that of nozzle loss coefficient. So, these are the numbers you would get and as you can see these numbers are quite different from that of axial flow turbine. So, we have done to solve two problems one on axial flow turbine, one on radial flow turbine and we have prescribed some problems for you to solve and as you can see you can get the feel of the numbers that the numbers corresponding to radial flow turbine are quite different from that of axial flow turbine. If you talk in terms of efficiency, if you talk in terms of rpm, you talk in terms of the mass flows the numbers of axial flow and radial flow turbine are quite different from each other and by solving the problems I hope you would get the feel of these numbers and that is very important for engineering people to get the feel of the numbers. We will leave the turbines with these problems and we will proceed in the next class, we will proceed on to the combustion chambers and that is what we will be doing in the next class. So, we have finished the turbine chapter and from the next class we will proceed on to the combustion chamber which actually spatially in a gas turbine you remember comes before the turbine. So, we will now look at how the hot gas is created that is applied to the turbine and that is what we will be doing in the next class.