 Hello and welcome to lecture number 36 of this lecture series on turbo machinery aerodynamics. The last class we had introduced a very new topic which is quite different from what we have discussed in some sense that is on radial flow turbines. Of course, we have already discussed its counterpart the centrifugal compressors, but in the context of turbines radial flow turbines operate in a quite different manner as compared to axial flow turbines. So, last lecture was exclusively devoted towards understanding of radial flow turbines, the basic working of radial flow turbines, the thermodynamics of radial flow turbine that is as the flow passes through a radial flow turbine, what in a thermodynamic sense happens to the flow as it passes through the different components of the turbine. We have discussed the different types of radial flow turbines which are possible, the two broad classes are the outward flow radial turbines and the inward flow radial turbines. We have seen that though the development the original development of radial flow turbines was in the out flow mode, very soon it was realized that the inflow mode or the inflow radial flow turbine is a much more efficient way of managing the flow and achieving work output than the out flow type of turbine. So, majority of the turbines which are being used currently are inward flow radial turbines. The radial turbines as a concept was originally developed towards meeting the hydraulic power requirement that is to convert hydraulic power into work output and then that continued for a very long time. In fact, that even continues till date and the hydroelectric power plants that we are aware of use one or more types of radial flow turbines. Of course, some of them also have axial flow types, but apparently radial flow turbines are very commonly used and the use of radial flow turbines for gas turbines was much later that it was also being considered as one of the options for use in a gas turbine engine. And the modern day gas turbine technology has restricted use for radial flow turbines it is primarily used in the smaller engine of smaller thrust class engines. And it is the main disadvantage when it comes to use of radial flow turbines for gas turbine applications is the fact that in order to achieve very high efficiencies and power output the turbine performance is a strong function of the inlet temperature. And in axial turbines we have already discussed that modern day gas turbines use different cooling technologies which can be used to which can actually permit us to use turbine inlet temperatures which are much higher than the material limits themselves. This is not it is still possible in radial flow turbines, but it is much more complicated to introduce and use some sort of cooling mechanisms in radial flow turbines. There are very complicated methods which are being proposed and probably being used in some types of engines. And this is one major disadvantage by radial turbines are not really used in gas turbine technology in gas turbine engines for larger thrust class engines because the turbine inlet temperature is sort of restricted. We saw that there are two distinct types of inward flow turbines the cantilever type and the 90 degree inward flow turbines. And we discussed in the previous class we devoted most of the time towards discussion on the 90 degree inward flow turbine because that is more commonly used. The cantilever type turbine is very similar to the impulse turbine that we have already discussed in the axial turbine context. And therefore, there are certain disadvantages associated with impulse turbine as we have already seen. And so the inward flow 90 degree inward flow turbine is like a reaction turbine. And in geometrical terms a 90 degree inward flow turbine looks exactly same exactly similar to a centrifugal compressor. But of course, just that the direction of flow and the rotation of the impeller or the rotor is exactly opposite in each cases. We also discussed about the governing equations which are used when we analyze the flow in these different components starting from the outlet flow the volute or the scroll. And then the nozzle blades and then it goes into the rotor or impeller. And then there is the exducer which forms the later part of the impeller it is counterpart and centrifugal compressor was the inducer. And then in typical turbine we may also have a diffuser downstream of the rotor to recover part of the kinetic energy which would otherwise be lost. And I think we also discussed two other aspects of radial flow turbines the efficiency and performance parameters. We define different forms of efficiency the total to static efficiency and how it is related to total efficiency and the work output and so on. We also discussed or spent quite some time discussing about the different loss parameters which are used in radial turbines like the nozzle flow coefficient and the rotor flow coefficient. We also looked at the incidence losses that is the flow entering the rotor need not necessarily be at zero incidence. And off design conditions the incidence angles could be quite high which leads to a substantial increase in the losses that is an additional component of loss that comes into picture when a turbine is operating in an off design condition. So, these are the different loss mechanisms that we discussed in little bit detail we did not of course spent lot of time discussing them because I think it is fairly out of the scope of this course to discuss the loss mechanisms and design optimization techniques in detail that is a vast subject on its own. So, which is why I decided not to spend too much time on that we will discuss some aspects of design and performance in the next class. But today's class I thought it is a good idea to have a tutorial session we shall be discussing a few problems which I will solve for you here. And then I also have a couple of exercise problems which I think you should be able to solve based on what we had discussed today as well as in the previous class. So, today's lecture is going to be a tutorial session and so let us take a look at the first tutorial problem that we have let us take a look at what the problem statement is. So, the problem number one that we have for today is the following the rotor of an inward flow radial turbine which is designed to operate at the normal condition is 23.76 centimeters in diameter and rotates at 38140 rpm. The design at the design point the absolute flow angle at the rotor entry is 72 degrees. The rotor mean exit diameter is one half of the rotor diameter and the relative velocity at the rotor exit is twice the relative velocity at the rotor inlet determine the specific work done. So, in this particular problem it is basically a very simple problem that of course the first problem that I normally solve is a very simple problem. We have some dimensions of the rotor we have the rotor diameter we have the rotational speed and the nozzle inlet angle. We also have been given that the rotor mean diameter is one half of the rotor exit diameter and the relative velocity at rotor exit is twice the relative velocity at the rotor inlet. So, based on this data that we have we need to find a specific work done. So, this is I would say very simple problem but again as I keep emphasizing every time I have a tutorial session is that we start solving a problem with the velocity triangles. So, let us construct the velocity triangles in this particular case and then we shall see and we shall proceed towards solving this problem because velocity triangle will help us understand what are the no parameters which are known to us and what are those parameters which we need to estimate and calculate. So, for a 90 degree IFR turbine we have the velocity triangles as shown here. We had already discussed this in the last class let me again just quickly explain the construction of a 90 degree IFR turbine. We have a volute or a scroll which sort of acts like a plenum chamber through which the flow enters a stagnation chamber here where the flow is stagnant and then that gets accelerated through the nozzle blades. So, these are the nozzle blades nozzle exit station is denoted by station 2 the flow then passes through the rotor blades or the impeller and exits at station 3 before going into the diffuser which is exit of the diffuser denoted by station 4. You can see that as I mentioned this is very similar to a centrifugal turbine a centrifugal compressor. In that case of course, the velocity the flow direction is the other way around it is reverse the flow actually proceeds in this direction. And instead of nozzle blades we have diffuser wings and therefore, these airfoil orientation will also be the reverse. Let us look at the velocity triangles for this case we normally have v 2 which is entering the rotor radially. So, the nozzle flow leaves relative velocity entering the nozzle is in the 90 degrees and the c 2 is at an angle of alpha 2 in this case it is given as 72 degrees. So, alpha 2 in this particular case is 72 degrees u 2 is the blade speed at the station that is station 2. Now, as the flow leaves the rotor the absolute velocity leaving the rotor is axial under design condition relative velocity is that an angle of beta 3 with the axial direction and that is v 3 and u 3 is the rotor speed at station 3. So, this is a typical inward flow radial turbine and the corresponding velocity triangles at the rotor inlet and rotor exit. So, based on the data that we have for this case we basically have the rotational speed we have the exit diameter. So, I think we should be able to find out u 2 and then subsequently we also have been given some ratio of the blade speed at the rotor exit to the mean diameter and the relative velocity at rotor inlet and exit. So, with this data we should be able to find the specific work done. Now, let me recall what we had discussed in the last class when we had derived a very general expression for a 90 degree I of R turbine where the specific work was if you recall a function of three distinct parameters. One is a difference between the blade speed at the inlet and outlet u 2 square minus u 3 square. The second term was a function of the relative velocity and the third term function of the absolute velocities. So, we are going to do exactly the same thing here to calculate the specific work done. Let us calculate these three individual components and then add up all of these and that gives us the specific work done. So, specific work done was 1 by 2 into 3 different terms u 2 square minus u 3 square plus w 2 v 2 square minus v 3 square and the third term was the absolute velocity. So, let us get these individual terms first add them up and then we get the specific work done. So, blade speed at the tip is u 2 which is basically pi d into n divided by 60 and so n has been given as 38, 140 rpm the diameter is given as 23.76 centimeter. So, it is 2.2376 this divided by 60. So, here u 2 comes out to be if you substitute these values we get u 2 as 474.5 meters per second. Now, this is where the velocity triangles come into picture that if we need to calculate v 2 if we have the velocity triangles in front of us it is very straight forward. It is ratio between v 2 and u 2 is related by tan alpha 2 or cot tangent alpha 2 cot alpha 2. So, let us look at the velocity triangle v 2 and u 2. So, tan alpha 2 is u 2 by v 2 and therefore, v 2 is u 2 into cot alpha 2 alpha 2 is given as 72 degrees and therefore, v 2 the relative velocity at rotor inlet is 154.17 meters per second. Similarly, c 2 is u 2 sin alpha 2 c 2 is this and this. So, sin alpha 2 is u 2 by c 2 and therefore, c 2 is u 2 sin alpha 2 and since alpha 2 is 70 degrees we get c 2 as 498.9 meters per second. Now, c 3 square that is the absolute velocity at the exit c 3 square is v 3 square minus u 3 square v 3 is given as twice of v 2 and that is 2 into 154.17 square and u 3 is related to u 2 because the diameter at u 3 station u 3 is half that of u 2 and therefore, u 3 is 0.5 times u 2. So, 2 into v 2 and 0.5 times u 2. So, what we get is that the square of this we basically get c 3 square c 3 square is 38786 meters square per second square. Similarly, let us find out the 3 individual components u 2 square minus u 3 square this is u 2 square into 1 minus 1 by 4 this is 1 by 2 square and so, it is 1 by 4 this should be 168863 meter square per second square. The second term is v 3 square minus v 2 square that is 3 into v 2 square because v 3 is twice of v 2. So, this is 71305 meter square second square per second square. The third term is c 2 square minus c 3 square we already know c 2 and we know c 3 square already. So, that difference is 210115 meter square per second square. So, these are the 3 individual components which contribute towards the specific work done. So, the next specific work done would be 1 by 2 times the sum of these 3 components. So, we add up all the 3 divide that by 2 we get the specific work done. Now, this is one way well probably the direct way of calculating specific work done. We can also approximate specific work done without having to undergo any of this, but of course, this is an approximate estimate of the specific work done that is simply equal to the square of the blade speed at the exit. So, delta w is u 2 square and why is it u 2 square? That is because the flow enters the rotor and leaves the rotor in the axial direction. So, c w 3 is 0 c w 2 will basically be equal to u 2 specific work done is u 2 c w 2 minus u 3 c w 3. The second term would become 0 the first term is equal to u 2 square. So, delta c w should also be equal to simply u 2 square. So, if we do that if we calculate work done in both ways let us see what happens. So, if we add up all the 3 different terms and divide by 2 we get the specific work done as 225 142 meter square per second square. And what we see is that the fractional contribution of each of these 3 terms the first term which is u square is 0.375 v square is 0.158 and c square is 0.467. That is you can see that there is a fair share for all these 3 components in the specific work done. So, we can calculate specific work done also by the second method which is basically because in this particular case beta 2 is 0 and alpha 3 is also 0 in which case delta w simply becomes u 2 square. So, if you simply square 474.5 the whole square you get 225 150 meter square per second square you can see that they are very close. Specific work done calculated in either of these ways should give you the correct answer both these methods give you identical specific work done. So, one way is to calculate the individual components and add up all of them which is a more general method because irrespective of how the velocity triangles are you could that is still valid. Whereas, delta w is equal to u square u 2 square is valid only if in this case like in this case the incidence is 0 the deviation is also 0 in which case you can directly calculate delta w. And one would get the same delta w excepting the round of errors which you see in both these calculations. So, this first problem as you can already as we have seen is a very simple problem which is involves simple application of mind to solving the velocity triangle to calculate the different components or constituents of the specific work done. One is the blade speed the other is relative velocity and the absolute velocity we add up all the 3 divided divide them by 2 we should get the specific work done. So, let us now move on to the next problem that I have for you and it is a slightly more involved problem. But of course, again I normally keep these problems limited to very fundamental aspects of the particular aspect that we design working on in this case it is the radial turbine. So, we are just looking at very fundamental thermodynamics of radial turbine and how we can apply some of these principles to calculate some parameters associated with radial flow turbines. Let us take a look at the second question we have for today. Problem statement number 2 is a radial flow real inflow turbine develops 60 kilowatts power when running at 60,000 rpm. The pressure ratio P 0 1 by P 3 of the turbine is 2 the inlet total temperature is 1200 Kelvin. The rotor has an inlet tip diameter of 12 centimeters and an exit tip diameter of 7.5 centimeters. The hub to tip ratio at the exit is 0.3 mass flow rate is 0.35 kg per second. The nozzle angle is 70 degrees and the rotor exit blade angle is 40 degrees. If the nozzle loss coefficient is 0.07 determine the total to static efficiency of the turbine and the rotor loss coefficient. So, here you can see of course, that the problem involves lot more data than what we had for the first case. We have the power input or power developed by the turbine that is 60 kilowatts the rotational speed 60,000 rpm the pressure ratio is 2 turbine inlet temperature is 1200 the dimensions of the rotor the tip diameter 12 centimeter exit tip diameter 7.5 hub to tip ratio 0.3 mass flow rate and the angles. And additionally the fact that the nozzle has a loss coefficient of 0.07. We need to find total to static efficiency and the rotor loss coefficient. So, this is the problem statement for this second question that we have as always we will first start with the velocity triangle. It is exactly the same as we have seen in the first problem. Nevertheless, let us just quickly look at the velocity triangles and understand what are the data that has been provided for us in this question and what is it that we need to find. So, in this case the velocity triangle again is the same as we have seen in the previous question. This angle is given to us as 70 degrees and the exit blade angle beta 3 is given as 40 degrees. We have the dimensions at station 2 and station 3 and hub to tip diameter ratio is also given to us. We have the power output and the rotational speed. So, quite a bit of data is given to us rotational speed is given means u 2 and u 3 are already known. Since angles are specified the other components can also be calculated. So, let us start calculating some of these parameters. The rotor tip rotational speed we can calculate from pi d n by 60 which is 377 meters per second and here d 2 is given in this question. It is given as 12 centimeters rotational speed is 60000 rpm. So, if we substitute pi and the rotational speed and the diameter divide that by 60 we get 377 meters per second. Now, from the velocity triangle we know that at the rotor inlet beta 2 is 0 and therefore, sin alpha 2 is simply u 2 by c 2 or c 2 is u 2 into cosecant alpha 2. Alpha 2 is given as 70 degrees and u 2 we have already calculated as 377 meters per second. Therefore, c 2 is equal to 377 into cosecant 70 degrees that is 401.185 meters per second. T 0 2 is given to us thermally temperature is 1200 Kelvin. Since c 2 is now calculated we can calculate t 2 static temperature that is t 0 2 minus c 2 square by 2 c p that should come out to be 1130 Kelvin. So, this is the preamble of the question that is we are solved some single parameters which may be required for solving the rest of the problem. So, first part of the question is to find the total to static efficiency and that is where we will make use of the pressure ratio that we have been given to we have been given as 2.0. From the pressure ratio we should be able to use that data to calculate the total to static efficiency and so that the next part of the question we going to solve is to find the stagnation temperature drop across the turbine and from the power output we can actually find the isentropic or the actual power output actual temperature drop across the turbine. If you take the ratio of the 2 we get the total to static efficiency. So, the stagnation temperature drop which is given by t 0 1 minus t 3 s is equal to t 0 1 into 1 minus t 3 s by t 0 1. We can convert this ratio into the pressure ratio because this ratio has been given as 1 by 2 that is 0.5 t 0 1 is known to us and therefore, this temperature drop is required for calculating the total to static efficiency because total to static efficiency is t 0 1 minus t 0 3 divided by t 0 1 minus t 3 s. So, t 0 1 is 1200 this multiplied by 1 minus 0.5 raise to gamma minus 1 by gamma you can calculate this and this should come out to be 190.92 Kelvin. Turbine power output as we know is m dot c p into t 0 1 minus t 0 3 power output is given as 60 kilo watts mass flow rate is 0.35 kgs per second c p we are assuming for gases as 1.148 therefore, t 0 1 minus t 0 3 is 149.33 Kelvin. So, if you simply take the ratio of this total to static efficiency is 149.33 divided by 190.92 and that is 0.782. So, total to static efficiency in this case is 0.782. So, that solves the first part of the question where we were required to calculate the total to static efficiency. Now, the next part of the question involves or requires us to calculate the rotor loss coefficient we have been given the nozzle loss coefficient as 0.07 and. So, we will need to make use of that rather complex formulae that we had derived in the last class if you remember there was a very long formulae which was relating the loss coefficient to the radius ratio and the static temperature ratios. So, we will make use of that formulae to calculate the nozzle well the rotor loss coefficient and that formulae involved both nozzle loss coefficient as well as rotor loss coefficient multiplied by the angles and so on. So, we know that in this case the radius ratio because this will be required in that formulae. Radish ratio R 3 by R 2 is basically the hub diameter at station 3 plus the shroud diameter divided by 2 that is the mean diameter divided by diameter at station 2 and this we know the hub to tip ratio is given as 0.4. So, we substitute zeta here and that is basically 0.4 in this question this multiplied by d 3 s plus d 3 s divided by 2 into d 2. So, if you substitute these diameters which have been given as well as the hub to tip ratio we can get the radius ratio as 0.406. Now, for efficiency this was the long expression I was mentioning total to static efficiency is 1 plus 1 by 2 into zeta n the nozzle loss coefficient t 3 by t 2 into cosecant square alpha 2 plus R 3 by R 2 square into zeta R which is the rotor loss coefficient into cosecant square beta 3 plus cot square beta 3 this whole raise to minus 1. Now, this though it looks rather complex formulae we have derived this expression right from the first principle. So, if you know if you can if you can actually relate the first principles formulae of total to static efficiency which is t naught 1 minus t naught 3 by t naught 1 minus t 3 s the numerator gets expressed in terms of the nozzle and the stator or the rotor loss coefficients numerator is expressed in terms of static temperature ratios and so on. So, from that we can actually derive this without much difficulty the numerator actually becomes C p times U into C p times C w delta C w and then delta C w we express in terms of the angles and so on. So, it is a very simple 2 3 step derivation from which we can derive this rather longish expression that we see here for total static efficiency. Now, in this expression we still have an unknown that is t 3 by t 2 t 2 we have already calculated, but we do not know the value of t 3 which also we can in fact calculate provided we know the exit stagnation temperature and the exit absolute velocity. So, if that is known we can actually we should be able to calculate because in this case beta 3 is given and from the velocity triangle since beta 3 is given we can and U 3 is known we can calculate C 3 and what about t naught 3 for t naught 3 the power output is given inlet stagnation temperature is given. So, we can calculate t naught 3 from there t naught 3 minus c 3 square by 2 C p will give us t 3 and then we can take ratio t 3 by t 2 that would be deriving the whole thing from the first principle or the whole thing can be expressed in a single definition term which also had I think mentioned in the last class which is basically in terms of some of these parameters which we know and a t 3 by t 2 is actually defined in terms of the velocities and the loss coefficients 1 minus U 2 square by 2 C p t 2 into 1 plus r 3 by r 2 the whole square multiplied by 1 plus zeta r cosecant square beta 3 minus 1 into cot square alpha 2 and. So, if you substitute for these values here the only unknown is zeta r. So, we should get 0.9396 minus 0.02187 zeta r. So, this if you substitute in this expression where zeta n is also known you get a quadratic equation and then we can solve that zeta r which is the rotor loss coefficient can be calculated as 0.62. So, this is one way of calculating zeta r the other way of course, is to calculate t 3 by t 2 using what I had mentioned t 2 we have already calculated we know what the value of t 2 is. For calculating t 3 it involves 2 3 steps 1 is to calculate the stagnation temperature at exit t naught 3 which can be calculated from the power expression power is is equal to mass flow rate into C p into delta t t naught 1 minus t naught 3. So, there all the three all the parameters are known except t 0 3 we can calculate stagnation temperature then to be able to calculate t 3 we also need to know the velocity at the exit that is c 3 and to calculate velocity at the exit c 3 we know beta 3 we also know u 3 and how do we know u 3 because the rotational speed is given to us and the diameter at the hub or at the exit of the rotor is also known. So, from there we can calculate u 3 since u 3 is known beta 3 is known we can calculate c 3 that is basically u 3 cot alpha 3 should be equal to c 3. So, once c 3 is also known the static temperature is at exit t 3 is equal to stagnation temperature t 0 3 minus c 3 square by 2 C p and so that is a rather easier way rather than this like longish formula as I just shown and then substitute for that in the expression and we should be able to calculate zeta zeta r which is the rotor loss coefficient nozzle loss coefficient is already known to us it is 0.07. So, what you can see is that the loss coefficient for the nozzle in this case has been given as 0.07 and for the rotor it is 0.62 and so I think I had also given some range of these values in the last class I had mentioned typically the rotor loss coefficients are on the higher side because of the fact that there are rotational effects coming to picture losses associated with the rotor are much more than the losses associated with the stationary component like a nozzle especially when the flow is accelerating. So, in this case the rotor loss coefficient is in fact close to one order magnitude higher than the nozzle loss coefficient. So, this solves our second problem which required us to calculate the total to static efficiency as well as the rotor loss coefficient. So, you can clearly see that this is a slightly more involved question in this of course I have taken the easier route of directly substituting this in the formulae what I would strongly urge you to do and I probably leave that as an exercise for you is to derive these equations from the first principles and not simply use the direct longish formulae it is very easy to derive the equations from the first principles. In the total to static efficiency definition term it is basically T 0 1 minus T 0 3 divided by T 0 1 minus T 3 s or let us express that in terms of enthalpy h 0 1 minus h 0 3 divided by h 0 1 minus h 3 s the denominator gets expressed in two separate forms one is to do with the nozzle loss coefficient second is the rotor loss coefficient numerator gets expressed in terms of mass flow rate C p and delta T and so on. So, from this you can actually denominator has nozzle and rotor loss coefficient term numerator is already known to us. So, this can be simplified and you can actually calculate the nozzle loss coefficient the rotor loss coefficient given the nozzle loss coefficient in a much more simpler less confusing manner than simply substituting them in the formulae. I have picked up this method because if you take up any text book you would normally see this kind of a method where they would refer to the derivation which was discussed earlier on like what I have done and the problem we just simply substitute plug in the values and calculate the corresponding efficiency and other terms required. Now, that brings us to the third question let us now proceed towards the third problem that we have for us to solve and then we will see how this problem is different from the previous ones. An inward flow turbine with 12 vanes is required to develop 230 kilowatts at an inlet stagnation temperature of 1050 Kelvin and a flow rate of 1 kilograms per second. Using the optimum design efficiency design method and assuming total to static efficiency of 0.81 determine the absolute and relative flow angles at the rotor inlet, part b the overall pressure ratio p 0 1 by p 3 and part c the rotor tip speed and the inlet absolute Mach number. So, this is a three part question where we have required to calculate three different aspects one is the angles absolute and relative flow angles the pressure ratio and of course, the rotor tip speed and the Mach number. And of course, you can see what is mentioned here is that we can assume optimum efficiency design method. I mentioned some preliminary aspects of this in the last class I would urge you to take up slightly more detailed reading of this in any of the text books that we have mentioned. You can take up pick up any book on the turbo machines and you will find a small section on optimum efficiency design methodology wherein a few formulae would basically be derived again from the first principles and you can see what it is basically trying to tell us. So, in this question it is a question which involves optimum efficiency design methodology which can be assumed and then we can we are required to calculate the angles and Mach number and so on. So, the first part of the question is to find the flow angles the relative and absolute flow angles. For optimum design it is known that the absolute angle at the nozzle exit that is alpha 2 is simply related to the number of blades and this comes from what is known as the white field width fields formulae which basically equates or which basically relates the alpha 2 to the number of plates. So, cos square alpha 2 is equal to 1 by n. So, that is where n is the number of blades. So, for optimum design it actually has been shown that by width field that cos square alpha 2 is 1 by n 1 by n where n is the number of blades. So, in this question we have 12 vanes and since we have been told to assume optimum design we can simply substitute the number of vanes here and calculate alpha 2. So, alpha 2 would be 73.22 degrees also as another consequence of the width field equation we have beta 2 is equal to 2 into 90 minus alpha 2. Beta 2 is the blade angle at the inlet of the rotor and that is equal to 2 into 90 minus alpha 2 and so this comes out to be 33.56 degrees. So, the next part of the question is to find the pressure ratio well basically the total to static efficiency and for which of course, we will use the total to static efficiency basically is T 0 1 minus T 0 3 by T 0 1 minus T 3 or T 3 as and from there we can express the denominator in terms of P 3 by P 0 1. And in this question we know that the power developed is given as 230 Kelvin inlet stagnation temperature is given mass flow rate is given and the total to static efficiency is given. So, we need to find the pressure ratio in this case. So, all we have to do is we just substitute the power output here which is m dot into C p into delta T C p is known assumed and the stagnation temperature is also can be calculated or it is given at the inlet. Since the efficiency is known we can calculate the pressure ratio P 3 by P 0 1 as 0.32165 the inverse of this is the turbine pressure ratio which is P 0 1 by P 3 and that is 3.109. So, this basically comes from the efficiency definition eta T s is T 0 1 minus T 0 3 divided by T 0 1 minus T 3 s the denominator gets expressed in terms of the pressure ratio from the isentropic relation. Numerator is simply C p times delta T and therefore, that is the power output of the turbine. Now, the third part of the question is to find the Mach number and then we also need to find the blade speed at the tip of the rotor. So, we will first find the absolute Mach number at the inlet for which we will find first the Mach number corresponding to stagnation conditions. So, m 0 2 this again is coming from the very basic definition for delta w m 0 2 square is equal to delta w by gamma minus 1 into 2 cos beta 2 by 1 plus cos beta 2. Now, where does this equation come from again I will urge you to try to derive this equation from the fundamentals delta w is h 0 1 minus h 0 3 which is m dot C p into T 0 1 minus T 0 3 and there the temperatures can actually be expressed in terms of the corresponding velocities. And therefore, we can express the Mach number in terms of the well part of the temperature gets expressed in terms of Mach number wherein these flow angles alpha 2 and beta 2 will also come into picture. So, from those fundamental equation you can actually derive the stagnation Mach number which is m 0 2 Mach number at the inlet of the rotor m 0 2 square is delta w by gamma minus 1 into 2 cos beta 2 by 1 plus cos beta 2. So, all the parameters on the right hand side are known to us we just substitute these different values and we get m 0 2 is 0.7389. The absolute Mach number which is basically based on the static conditions m 2 is related to m 0 2 as we know. So, m 2 square is m 0 2 square by 1 plus gamma minus 1 by 2 m 0 2 square. This again follows from the isentropic relations we have already seen the relation between stagnation temperature to static temperature, stagnation pressure to static pressure and so on. We can also relate the corresponding Mach numbers in this way. So, since we have calculated m 0 2 we substitute that here and then we get the static Mach number based on static conditions that the absolute Mach number as 0.775. So, this is again this can be calculated in multiple ways. The other way which I suggest you can try to calculate would be to take the ratio of the absolute velocity that is C 2 divided by square root of gamma R T 2. Stagnation temperature is known at the inlet and to find static temperature we need T naught minus C square by 2 C p. So, you basically need to calculate C square at the inlet and to calculate C square we will of course need the blade speed because you need to calculate because blade angle is known and so if you know the blade speed at the tip that is U 2 you can solve C 2 and then therefore calculate Mach number. In this case of course we are doing it the other way around we are calculating Mach number first and then now we are going to calculate the blade speed at the tip but it can also be done the other way around. So, let us now calculate the blade speed at the exit of the nozzle or that is the entry rotor entry tip of the rotor delta W by C p T 0 1 as we know is equal to gamma minus 1 by 2 gamma minus 1 into cos beta 2 U 2 square by A 0 1 square. So, here A 0 1 is speed of sound based on stagnation temperature square root of gamma R T 0 1. This can be calculated as 633.8 meters per second assuming T 0 1 is equal to T 0 2. Since in this case we know the power output and stagnation temperature and beta 2 we can simply substitute for these values and then we can calculate the blade speed at the tip U 2 and that comes out be 538.1 meters per second. So, the other approach to calculate Mach number would be to actually calculate the blade speed first and then since blade speed is known and alpha 2 is known you can calculate C 2 and from C 2 you can calculate the static temperature T 2 T 0 2 minus C 2 square by 2 C p and therefore, Mach number would be C 2 by square root of gamma R T 2. So, that is another way of calculating the Mach number of we have calculated that in a slightly different way. So, that completes the third problem that we had set aside for today's tutorial. So, I have one more problem to solve which is a very simple problem where basically not involving any calculation, but this is just to compare performance or operation of two different types of turbines that we have discussed about. One is the axial turbine which we had rather detailed discussion several lectures earlier on and of course, the radial turbine. So, let us compare under certain given operating conditions how the work output of these two different turbines can be calculated and how do they compare with each other. So, the fourth problem statement is the following. Compare the specific power output of axial and radial turbines in the following cases. Axial turbine in this case is given as alpha 2 is equal to beta 3 is 60 degrees and alpha 3 is equal to beta 2 is 0 degrees. For the radial turbine alpha 2 is given as 60 degrees and beta 3 and alpha 3 beta 2 all of them are 0. If the rotational speed is the same in both these cases, we are required to calculate the specific power output. So, what we can see is that immediately for the axial turbine we have we can see that alpha 2 is equal to beta 3 and alpha 3 is equal to beta 2 immediately tells us that this is a 50 percent reaction turbine which means that the velocity triangles would be symmetrical. And for the radial turbine we have already seen the velocity triangle which is for the nominal operation condition where alpha 2 is given beta 2 is 0 alpha 3 and beta 3 are respectively 0 at the exit. Let us take a look at the velocity triangles first and so here we have the velocity triangle for both these cases the axial as well as the radial turbines. Let us look at the axial turbine first velocity triangle at the inlet is given by this alpha 2 is given as 60 degrees and so alpha 2 at 60 degrees and that is equal to beta 3 that is also 60 degrees and beta 2 is 0 as you can see and alpha 2 is 0 that is C 3 makes 0 degrees with the axial direction. And since it is an axial turbine we will assume u 2 is equal to u 3 is equal to u blade speed it is for the same circumferential radial location. So, u 3 and u 2 are the same that is equal to u and these are the corresponding velocities C 2 at the inlet the absolute velocity v 2 is the relative and at the rotor exit we have C 3 and correspondingly v 3 which is the relative velocity. For the radial turbine we have been given that alpha 2 is again 60 degrees and beta 3 is also 60 degrees and this is a typical velocity triangle for an inward flow radial turbine. Since alpha 2 is 60 degrees we have C 2 which is at 60 degrees to the radial location here v 2 is the relative velocity which is entering in the radial direction u 2 and u 3 are not the same they are different in this case. And at the exit of the rotor we have beta 3 which is again 60 degrees and the axial velocity which is C 3 and C A 3 where they are the same v 3 is the relative velocity at the exit of the rotor. So, for the axial turbine we have alpha 2 and beta 3 as 60 degrees alpha 3 and beta 2 as 0 degrees specific work for this case in this 50 percent reaction case would be u times delta C w u times C w 2 and C w 3 here as you can see C w 3 would be equal to 0 for the axial turbine because the flow is leaving in the axial direction. So, C w 3 is 0 and C w 2 is equal to u 2 C w 2 is the tangential component of C 2 which is equal to u 2 and therefore, the specific work done for the axial turbine would be simply u square. For the radial turbine alpha 2 and beta 3 are 60 degrees beta 2 and alpha 3 are equal to 0 and specific work done is u 2 C w 2 minus u 3 C w 3. In this case of course, C w 3 is again 0 like in the case of axial turbine and C w 2 is the inlet is equal to u 2 and of course, at the exit u 2 and u 3 are not the same, but since C w 3 is 0 the radial turbine also develops a work which is equal to u square in this case it is u 2 square. So, specific work done in this particular case one of them is for an axial turbine and the other is for radial turbine for given these conditions for the same rotational speed both of these turbines generate the same work output. They are both functions of square of the blade speed. So, in this specific case that our example that we have looked at because of the conditions that has been specified to us for the same blade angles and rotational speeds both these turbines generate the same work output. So, this just to give you an idea of how the work done can be calculated for different types of turbine configurations of course, we have seen axial turbines in greater detail including a tutorial session earlier on and this is this was just to make a comparison between the work output of these two different types of turbine configurations. So, that completes the fourth problem as well that we have solved today and what I have for you are two exercise problems which I would leave it for you to solve and of course, I had also left a few exercises in between in couple of problems where I had requested that you should try to solve it in a different way. The method I had solved is one of the ways of solving the problem. You can also attempt to solve the problem in a different way for which I had given some hints. So, I suggest that you would also solve those problems using the other alternative approach that I had suggested. So, let us take a look at the first exercise problem that I have a small inward radial flow gas turbine comprising a ring of nozzle blades radial vane rotor and an axial diffuser operates at nominal design point with the total to total efficiency of 0.9. At the turbine entry the stagnation pressure and temperature of the gas is 400 kilo Pascal and 1140 Kelvin respectively. The flow leaving the turbine is diffuse to a pressure of 100 kilo Pascal and has negligible final velocity. Given that the flow is just choked at the nozzle exit determine the impeller peripheral speed and the flow outlet angle from the nozzles. So, this question has an additional component that is the diffuser and it is given that at the exit of the diffuser the flow has negligible velocity. So, you can assume that the flow exits the diffuser with almost 0 velocity and the flow at the nozzle exit is just choked. Which means that Mach number at nozzle exit is 1 velocity there would be equal to square root of gamma R T and stagnation temperature is given to us and. So, that should help you in finding out the parameters at the rotor entry and since rotor exit conditions are sort of fixed you can also calculate the conditions at the rotor exit using the fact that the diffusion pressure is given and the fact that the velocity at the exit is close to 0. So, in this case the tip speed comes out to be 586 meters per second and the angle alpha 2 is 73.75 degrees. Now, the second question exercise problem is if the mass flow rate of the gas through the turbine given in this problem the previous problem is 3.1 kgs per second the ratio of rotor axial width to rotor tip radius is 0.1 and the nozzle isentropic velocity ratio is 0.96. Assuming that the space between the nozzle exit and rotor entry is negligible and ignoring the effects of blade blockage determine the static pressure and static temperature at the nozzle exit the rotor tip diameter and rotational speed the power transmitted assuming a mechanical efficiency of 93.5 percent. So, we need to use part of data which is given in the first question to be able to solve this question as well. So, answer to part 1 is the pressure static pressure is 205.8 kilo Pascal temperature is 977 Kelvin the rotor tip diameter is 125.4 millimeters and rotational speed is 89200 rpm power transmitted with this mechanical efficiency would come out to be 1 megawatts. So, these are two exercise problems that I have for you you can solve this based on what we have discussed in the last couple of lectures including today's. And I would also suggest that you solve couple of those problems which we solved in today's lecture that is problem number 2 and 3 in a different approach from what we have solved in the tutorial. So, I would suggest that you also solve those problems using a slightly different approach. So, we would have just one more lecture on radial turbines where we would discuss some aspects of performance and some preliminary design aspects related to radial flow turbines. So, these we will take up in the next class which would be lecture number 37.