 Hello, and welcome to lecture number 33 of this lecture series on turbo machinery aerodynamics. As promised in the last lecture, we shall be having a tutorial session in today's lecture. But before I start the tutorial, let me quickly recap what we have been discussing in the last two lectures. As you know that last two lectures have been devoted exclusively for centrifugal compressors. So, during the first lecture on this series that was on lecture number 31, where we introduced the aspect of centrifugal compressors, we discussed in detail the thermodynamics of centrifugal compressors. And we also discussed in some detail about why is it that centrifugal compressors have certain amounts of benefits in some cases and as well as the fact that there are lot of disadvantages associated with centrifugal compressors and that is primarily the reason why these compressors are not used in large sized engines. So, the application of centrifugal compressors are primarily at this moment limited to smaller sized or smaller thrust class engines and because axial compressors have inherent advantages over centrifugal compressors, which makes them more suitable for larger sized engines. But centrifugal compressors continue to be used in smaller engines and of course, they were used popularly in the early days of development of the jet engines when axial compression system was not that developed in those days. And then we also discussed about the different components, which constitute a centrifugal compressor like the inlet part of the centrifugal compressor, the inducer. Induser is one which guides the flow into the impeller to ensure that there is smooth entry of the flow into the impeller. Then the impeller itself, which is the rotor of the centrifugal compressor. Induser is often attached to the impeller and it forms the initial part of the impeller itself. Of course, in some of the older generation engines and compressors, inducer was sometimes kept as a separate component and not necessarily a part of the impeller itself. Impeller flow gets discharged into a vaneless space, which is where the diffusion continues from the impeller, it then continues in the vaneless space. We have seen the physics behind that and then the flow proceeds into the diffuser, which could be either vane type diffuser or it could be pipe diffuser, channel diffuser and so on. From the diffuser, the flow is collected through what are known as collectors or volute and then it is discharged from the compressor. So, these are different components, which constitute a centrifugal compressor. We have seen the flow as it passes through these different components, how we can calculate different parameters associated with the flow. So, these were some of the things we had discussed in the first lecture, which was on during lecture number 31. In the previous lecture, we continued our discussion on this on little more advanced concepts like the Coriolis acceleration and what its significance is in reference to centrifugal compressors. We have seen that Coriolis acceleration basically leads to a tangential variation in the relative velocity and at the exit of the impeller, this would mean that there is what is known as a slip. Slip is basically referring to a difference between the tangential component of the absolute velocity to the tip blade speed that is C w 2 by u 2 and so slip factor basically affects the performance in terms of pressure rise of the centrifugal compressor. We have seen that slip factor is a strong function of the number of blades. Therefore, there are empirical correlations, which from which one can calculate slip factor for different types of impellers and subsequently, we discussed in detail about the performance characteristics of centrifugal compressor. We have seen that it is very similar to that of axial compressors and we spend some time discussing about the choking aspect of centrifugal compressor and also the fact that choking is different in different components that is the way choking is calculated in the impeller or the that is the rotating components or the stationary component like the inlet or the diffuser is quite different and also the fact that one can continue to operate the engine at a higher mass flow than the choking mass flow, if we can increase the speed of the impeller and also the fact that this does not lead to choking of the other components that is as long as all the other components are not choked for a given operating condition, if the compressor is operating under choked condition, if we increase the rotational speed we can actually operate at a slightly higher mass flow rates at least theoretically. These were the different aspects that we had discussed in the last two lectures and therefore, it is about time that we have a tutorial session and I think I also mentioned why we are not emphasizing too much on the centrifugal or the radial flow machines like centrifugal compressors or you will see shortly about radial turbines is the fact that these components or these types of turbo machines though have been used extensively in the past in the current day scenario their application is very much limited and most of the modern day jet engines that we are aware of the larger sized ones definitely use axial flow turbo machines like the compressors and turbines. So, today's lecture we will devote towards centrifugal compressors on basically trying to solve some problems on centrifugal compressors. Let us look at basically have a tutorial session on centrifugal compressors. So, we will have a tutorial on centrifugal compressors. Let us take a look at the first problem that we have today. So, the first problem statement is the following at the inlet of a centrifugal compressor I the relative Mach number is to be limited to 0.97 the hub to tip radius ratio of the inducer is 0.4 the I tip diameter is 20 centimeter and if the inlet velocity is axial determine part A the maximum mass flow rate for a rotational speed of 29 160 rpm part B the blade angle at the inducer tip for this mass flow the inlet conditions can be taken as 101.3 kilo Pascal's and 288 Kelvin. So, let us read this question little more carefully what is basically the problem statement referring to it tells us that there is a relative Mach number at the inlet I and that is limited to 0.97 that is we have to keep the Mach number limited to 0.97 and not let it attain supersonic speech the half to tip ratio is given as 0.4 the I tip diameter is given as 20 centimeter then we have the blade rotational speed we need to calculate maximum mass flow and the blade angle. So, let us take a look at the schematic of this particular problem we have this half to tip ratio that is been given to us the radius ratio is 0.4 which means that R H by R T is 0.4 the inlet velocity is given as axial. So, C 1 is axial and then the blade angle beta 1 is one of the parameters we need to calculate since u 1 is also specified because we know the tip diameter and the rotational speed. So, from there we can calculate the blade angle beta 1. So, this basically is a problem which refers to a typical centrifugal compressor problem which has the flow entering the inducer axially in this case the flow is indeed axial and that is one of the parameters which has been given rotational speed is given some of the geometrical dimensions with reference to the inducer is also given that the tip diameter the half to tip ratio etcetera is given to us. And one of the important statements in the problem is that we need to ensure that the Mach number is limited to 0.97 at the inlet actually start solving the problem from there onwards. So, we know that the Mach number is to be limited to that and then from there we can calculate let us say the static temperature because inlet stagnation temperature is given the inlet velocity is basically can be calculated. So, if you look at the rotational speed we have been given that the rotational speed is 29 160 rpm diameter is 0.2 20 centimeters. And so, we have pi d n by 60 this is 305.36 meters per second. Now, from the velocity triangle we can see that the relative Mach number is basically relative velocity by square root of gamma r t 1 and at the inlet v 1 is equal to square root of c 1 square plus u 1 square this divided by square root of gamma r t 1. And so, we also know that t 1 is equal to t 0 1 minus c 1 square by 2 c p and t 0 1 is given as 288 c 1 of course, is not known c p we will assume for air as 1005 joules per kilogram Kelvin. And therefore, 2 c p is 2010. So, here we have 2 equations one is for Mach number and the other is for static temperature. So, let us substitute this in this equation relative Mach number becomes square root of c 1 square plus u 1 square divided by square root of gamma r into the temperature 288 minus c 1 square by 2010 right hand side is involves an unknown that is c 1 and left hand side is already given that we have to limit the Mach number to 0.97. So, if you substitute all the values we have 0.97 square is equal to c 1 square plus 305.63 square divided by gamma r into 288 that is 115718.4 minus 0.2 c 1 square. So, if you simplify this it is a simple quadratic equation which we can simplify and then we can get c 1 is 114.62 meters per second. So, if you refer to the velocity triangle we have basically determined c 1 and since u 1 is known you can now calculate beta 1 because it is basically the time of u 1 and c 1 of course, the first part of the question is also to find the mass flow rate. So, let us calculate the mass flow rate for calculating mass flow rate we need the density axial velocity is known we also need the area. Now, the static temperature you can calculate here because the stagnation temperature is given c 1 we have just now calculated and therefore, static temperature would be 288 minus c 1 square by 2 c p c 1 square we have calculated and therefore, it becomes 281.46. And then since no efficiency has actually been specified we can use isentropic relations here to calculate the corresponding static pressure because stagnation pressure has been specified stagnation temperature and static temperature are known. So, if you substitute these values here we can calculate the static pressure p 1 which is 93.48 kilo Pascal's. Now, once we calculate the pressure and temperature we can calculate the density which is p 1 by R T 1 and that is 1.157 kilo grams per meter cube. So, this is one of the parameters which we need for calculating mass flow rate the other parameter is the area of passage of the mass flow rate and that is the annulus area at the inlet for which we have been given the tip diameter and the hub to tip ratio. So, the annulus area would be pi d square by 4 into 1 minus R H by R T. So, that is 0.0264 meter square. So, this is the annulus area through which the mass flow rate passes and we have calculated density. Now, the third parameter for calculating mass flow rate is the velocity and for mass flow rate we need to know the axial velocity, but at the inlet we know that the absolute velocity itself is axial and therefore, C A 1 should be equal to C 1. So, mass flow rate is quite straight forward now we just multiply density annulus area at inlet and the axial velocity. So, rho 1 A 1 and C 1 the product of these 3 would give us the mass flow rate and so we multiply the density area and absolute velocity we get the mass flow rate. Then the second parameter is the blade angle. So, how does one calculate the blade angle? Blade angle at the inlet is very straight forward again because we have already calculated C 1 and we know the blade speed u. So, tan inverse of C by u C 1 by u 1 would give us the angle at the inlet that is beta 1. So, if you substitute those values here we get mass flow rate as rho 1 A 1 C 1 which is 1.157 into 0.0264 which is the annulus area C 1 is 114.62 meter per second. So, the product of all the 3 will give us the mass flow rate which comes out to be 3.5 kilograms per second blade angle at the inlet is at the tip is tan beta 1 which is C 1 by u 1. Therefore, beta 1 is tan inverse C 1 by u 1 which comes out to be 20.57 degrees. So, these are two parameters that we have calculated the mass flow rate which was the first part of the question. Second part was for the same mass flow what is the blade angle and so we have also calculated the blade angle for this question. I think I have mentioned this several times in the past that the key to solving problems to do with turbo machines the fundamental problems to do with turbo machines is to get the velocity triangles. So, even for a centrifugal compressor as you have seen velocity triangle is the starting point for any of this analysis. So, you start solving the problem once you get the velocity triangles. So, if the velocity triangles can be set right then solving the problem is quite easy because you know what to calculate and how to calculate them. So, I would urge you to keep this in mind whenever you are solving a problem irrespective of whether it is axial compressors axial turbines centrifugal or radial turbines. Whatever be the problem that you are solving it is very important that you understand the significance of velocity triangles and that is why you need to understand the physics very well to be able to construct the velocity triangle and that would basically be the starting point for solving any of these problems. So, let us move on to the next problem now and let us see what this problem statement is. Second problem statement is that a centrifugal compressor has a pressure ratio of 4 is to 1 with an isentropic efficiency of 80 percent when running at 15000 rpm and inducing air at 293 Kelvin curved vanes at the inlet give the air a pre whirl of 25 degrees to the axial direction at all radii. The tip diameter of the eye of the impeller is 250 millimeters the absolute velocity at inlet is 150 meters per second and the impeller diameter is 600 millimeters calculate the slip factor. So, this is in fact a follow up question for this will also will be given a little later we will also solve that problem which is quite similar to this, but with a small twist. So, in this case the question states that we have a centrifugal compressor which is developing a pressure ratio of 4 is to 1 with a certain efficiency and rotational speed what is to be noted here is that the air is no longer entering axially. Unlike the previous question where it was specifically mentioned that the inlet air is axial here it is not axial it is coming with a pre whirl that is a pre whirl is like guide vane ahead of the inducer where the guide vane set the flow at a certain angle in this case the angle has been set at 25 degrees. And then we have been given some dimensions of the impeller and some speeds. So, based on this we need to calculate the slip factor. So, slip factor as we have learnt in the previous lecture is basically the ratio of the tangential absolute velocity at the impeller exit divided by the corresponding blade speed at the same location. So, it is C w 2 by u 2 which is the slip factor we are required to find this. Now, since our rotational speed is given and the diameter of the impeller is given finding the blade speed at the tip is very easy because u 2 is pi d 2 n by 60 and d n both are given to us. And so you can calculate the impeller tip diameter tip speed, but how do you calculate C w 2? This of course, will require us to solve the velocity triangle first at the inlet and subsequently at the exit as well to be able to solve this problem. So, let us look at the velocity triangle at the inlet of the inducer. So, these are the inducer veins which are shown and you can see that the flow is not entering the flow in inducer in an axial direction it is in fact entering at an angle of 25 degrees. That means that C 1 is not axial C 1 itself has a tangential components C w 1 then the inlet relative velocity is v 1 and the blade speed at the inlet is u 1. So, this is which means that there would be a set of veins which will set this flow angle of 25 degrees to ensure that the absolute velocity enters the inducer in this direction and that is called a pre whirl. You can see this word mentioned here it is called pre whirling. Pre whirl means that there is a whirl component which is basically the tangential component as of absolute velocity and that is why it is called a pre whirl. In the absence of pre whirl the flow would have entered axially, but that may not really be an ideal condition for the inducer because the relative velocity might actually enter the inducer at a different angle of course, this has been designed for a pre whirl. So, let us begin to solve this problem we have the pressure ratio given to us and therefore, we can calculate the exit stagnation temperature exit stagnation temperature is equal to T 0 1 times the pressure ratio raise to gamma minus 1 by gamma and therefore, we get 293 into 4 raise to 1.4 minus 1 by 1.4 that is 435.56 Kelvin. So, the isentropic temperature rise would be 435.56 minus T 0 1 that is 293 we get 142.56 Kelvin, but the actual temperature rise is isentropic temperature rise divided by the efficiency which in this case is specified as 80 percent and therefore, we get delta T naught is equal to 178.2 Kelvin which is 142.56 divided by 0.8. Therefore, from the temperature rise we can calculate the work done per unit mass and work done is can be calculated in two ways as you have seen in the past. One is by using the temperature rise and the other method is by using velocity triangles that is you delta u C w will basically give us the work done per unit mass. And so here we know since we know the temperature rise we can calculate the work done per unit mass which is basically the enthalpy rise in the compressor and that is C p times delta T naught. Delta T naught we have calculated as 178 this multiplied by C p will give us the work done per unit mass. So, work done here would be C p which is 1.005 into delta T naught delta T naught is 178.2 therefore, we get 179 kilo joules per kilogram. Now at the exit we can now calculate some parameters at the exit because our aim is to calculate the slip factor at the exit and but of course, before that we need to calculate C w 1 because that is required for calculating the exit parameters as well. Now the inlet of the i we have u 1 is pi d n by 60 that is pi into the tip of the i of the impeller is 0.25 meters into 15000 divided by 60 and that comes out to be 196.25 meters per second. C w 1 is C 1 times sin 25 C 1 is given to us and therefore, and also the angle is given alpha C 1 sin alpha 1 therefore, that comes out to be 63.4 meters per second. Now peripheral velocity at the tip of the impeller u 2 would be pi into capital D which is the tip diameter of the impeller that is given as 600 millimeters multiplied by n by 60. So, pi into 0.6 into 15000 divided by 60 this is 471.2 meters per second. So, here we have calculated the conditions at the inlet which refers to the C w 1 and why we are calculating C w 1 is because we have just now calculated the work done per unit mass. And work done per unit mass is also equal to u 1 u 2 C w 2 minus u 1 C w 1 and from there u 1 and u 2 are known C w 1 we have just now calculated and therefore, we can calculate C w 2. Once you calculate C w 2 ratio of C w 2 to the blade speed at the tip that is u 2 will give us the slip factor. So, let us now calculate C w 2 which is the world velocity at the tip of the impeller and since the power work done per unit mass is also equal to u 2 C w 2 minus u 1 C w 1 and this is equal to work done we have just calculated earlier 179 kilo joules per kilogram. This is equal to u 2 times C w 2 that is 471.24 into C w 2 minus u 1 C w 1 that is 196.35 into 63.4. So, C w 2 we can calculate as 406.27 and therefore, we calculate slip factor as ratio of C w 2 to u 2 and therefore, 406.27 divided by 471.24 that is 0.862. So, here we get a slip factor of 0.86 what are the implications of this? Well the implications are depends upon the kind of application we have, but the basic implication of having a slip factor much lower than 1 is that it leads to lower and lower pressure ratios that is higher the slip factor lower is the pressure ratio for which it is actually been designed for. This is because the tangential velocity at the exit that is basically the solid velocity at the exit is directly related to the pressure ratio. You can actually derive an expression which relates the pressure rise P 0 2 by P 0 1 in terms of C w and u 1. So, therefore, there is a direct correlation between that. So, lower the slip factor which means lower is C w 2 in comparison to u 2 and therefore, that will affect the pressure rise achieved in a centrifugal compressor and that is why modern day designers would like to maximize or increase the value of slip factor to keep it as close as possible to 1. We have seen the strong dependence of slip factor on the number of blades and that is one of the key optimization challenges because one can keep increasing number of blades, but then the problem with that would be the fact that increasing number of blades also leads to an increase in these skin friction losses and therefore, that is going to affect the efficiency in some way. The designer does not want to have a poor efficiency with a higher pressure ratio that does not make sense. So, one needs to have a choice of will basically a mix of high efficiency as well as higher pressure ratio and that requires a very intelligent way of trying to optimize this case where we have one on one hand one option of increasing number of blades with the risk of lower efficiency because of losses. Other option is to reduce number of blades and have higher efficiency possibly higher efficiency because of lower frictional losses, but one may end up with a lower pressure ratio. So, there is a trade off required to attain an optimum condition. So, let us now again look at another problem, third problem also involves slip. Let us take a look at what the problem statement is. The problem number 3 states that air at a stagnation temperature of 22 degree Celsius enters the impeller of a centrifugal compressor in the axial direction. The rotor which has 17 radial vanes rotates at 15,000 rpm. The stagnation pressure ratio between the diffuser outlet and the impeller inlet is 4.2 and the total to total efficiency is 0.83 percent or 0.83. Determine the impeller tip radius assuming that the air density at the impeller outlet is 2 kgs per meter cube and the axial width at the entrance to the diffuser is 11 millimeters. You also need to determine the absolute Mach number at this point. We assume that the slip factor is 1 minus 2 by n where n is the number of blades. So, we have been permitted to use the standard slip factor which states that the slip factor is equal to 1 minus 2 by n n being the number of vanes. So, in this question we have been given the number of blades here 17 radial vanes rotational speed 15,000 rpm and the stagnation pressure ratio 4.2 total to total efficiency 83 percent and then the density at the impeller outlet and the axial width at the entrance of the diffuser. So, these are the parameters which are given to us based on which we need to calculate two parameters. One is the impeller tip radius and the second is we also need to calculate the Mach number at the exit of the impeller. So, we need to of course, calculate the absolute Mach number. So, this is a question which is very similar to one of the questions we have solved earlier. So, I will skip the velocity triangle part leaving that to you to figure out the velocity triangle and try and construct the velocity triangle for that case. So, this question involves requires us to calculate two parameters one of course, the hint given to us is that we can calculate slip factor using the Steinitz formula we have we have been given the number of blades. So, we can actually calculate the slip factor from there and then we know the pressure ratio and total to total efficiency and. So, we can actually calculate the power required from what data has been given to us. The other key information is that the flow enters the impeller in the axial direction which means that there is no tangential component to the absolute velocity at the inlet. So, C w 1 should be 0 and so you can calculate work done based on just u 2 times C w 2. So, the specific work required is u 2 C w 2 minus u 1 C w 1 C w 1 is 0 and therefore, work is u 2 C w 2 which is also equal to sigma times u 2 square because sigma is C w 2 by u 2 and. So, we can express work done in terms of sigma which is the slip factor and the blade speed at the tip of the impeller u 2. So, sigma u 2 square would be the work done specific work done. So, we can now express efficiency in terms of well u 2 in terms of efficiency and pressure ratio. This comes from the total to total efficiency we have already defined that earlier on in the context of a turbine. So, similar aspect we can also apply for this case. So, when we use the efficiency definition we have the outlet temperature T 0 3 minus T 0 1 T 0 3 being the isentropic temperature minus T 0 1 divided by T 0 3 minus T 0 1 and then the pressure ratio is given to us. So, the numerator gets converted or can be expressed in terms of the pressure ratio. Pressure ratio has been given to us as 4.2 and the denominator which is T 3 minus T 0 1 is basically the work done divided by specific heats. So, you can express that in terms of w times w divided by C p w is sigma into u 2 square and therefore, we can express u 2 in terms of these parameters which are known to us the inlet stagnation temperature sigma the temperature and pressure ratio as well as the efficiency. So, if you express u 2 in terms of efficiency and pressure ratio we have u 2 square is C p T 0 1 multiplied by pi c raise to gamma minus 1 by gamma minus 1 divided by sigma into efficiency. Efficiency is given as 83 percent and sigma is given can be calculated based on the number of blades. So, sigma is 1 minus 2 by n, n is 17 we have 17 number of vanes of blades. So, sigma the slip factor is 0.8824. So, we know the temperature you know pressure ratio and the efficiency if you substitute that we get the blade speed as 452 meters per second. And then omega because the speed is given to us as 15000 rpm based on that we can now calculate the tip radius because tip radius is u 2 or the tip speed is basically equal to omega into r. Therefore, r t is equal to u 2 by omega and omega here is the rotational speed in radians per second 15000 into pi by 60 we get 1570 radians per second. So, if you substitute for those two 452 divided by 1570 we get the tip radius as 0.288 meters. So, this is the first part of the question where we are required to find tip diameter of the impeller which in this case comes out to be 288 millimeters or 0.288 meters. The second part of the question is to find the Mach number well the absolute Mach number at the impeller exit which means that we have to take the ratio of C 2 to the speed of sound at that location that is A 2. So, C 2 by A 2 will give us the Mach number at station 2 which is impeller exit. And so we need to now find out the absolute Mach number we also need to know the temperature static temperature at the exit of the impeller to be able to calculate the Mach number. Now, at the exit of the impeller we have C 2 which is root of C w 2 square plus C r 2 square where C r is the radial velocity the absolute component of radial velocity. And how do we calculate C r 2? C r 2 would basically be coming from the mass flow rate mass flow rate divided by the density times the annulus area. Now, in this case we have mass flow rate which has been given to us as 2 kilograms per second density is also given as 2 kg per meter cube tip radius we have just now calculated 288 millimeters or 0.288 meters. And the axial width is given to us in the question as 11 millimeters. So, this is 0.011. So, C r 2 which is the radial velocity at the exit of the impeller is mass flow rate divided by density times the annulus area 2 pi into tip radius times the axial width which gives us the annular area this comes out to be 50.3 meters per second. Similarly, C w 2 we can calculate because we know u 2 we also know the slip factor sigma product of those 2 we would give us the tangential velocity at the exit C w 2. So, this is 400 meters per second. So, C 2 is equal to square root of 50.3 square plus 400 square that is 402.5 meters per second. So, for calculating Mach number we now have the absolute velocity we now need to calculate the static temperature as well. So, for calculating the static temperature we will make use of the rothalpy we have already discussed about that earlier on. So, we will make use of that concept here the conservation of rothalpy in the impeller and based on that we will calculate the static temperature. So, we know that h 0 2 which is stagnation temperature at the exit of the impeller would be equal to inlet stagnation temperature plus the work done that is w c. So, h 0 1 plus w c will be h 0 2 and therefore, h 0 1 plus sigma u 2 square because w c we know is in this case is sigma u 2 square and therefore, h 2 would be equal to h 0 1 plus sigma u 2 square minus half c 2 square because h 2 plus half c 2 square is h 0 2. This we will convert in terms of temperatures since h 2 is equal to h 0 1 plus sigma u 2 square minus half c 2 square correspondingly in terms of temperature we get t 2 is equal to t 0 1 plus sigma u 2 square minus half c 2 square divided by c p. So, we substitute all these values and we get temperature static temperature at the impeller exit as 394.5 Kelvin. Therefore, the Mach number at the exit of the impeller is 402 which is the absolute velocity divided by square root of 1.4 into R 287 into 394.5 Mach number comes out to be 1.01. So, we can see that the absolute Mach number is just about sonic it has just crossed the sonic Mach number and in fact, in the relative frame of reference the Mach number can in fact be higher than what you have calculated for the absolute case here. So, the third problem we just now solved concerned about was basically about calculating firstly the Mach number we have seen how to calculate the Mach number as well as how do you calculate the based on the parameters like the slip factor in this case to be approximated as 1 minus 2 by n. We can also calculate the mass flow rate as we have done in this question. Now, let us take up one more problem which is in some sense identical to what we have solved in the second question and partly also identical to what we have just now solved for the third question. So, we will require understanding of how those two problems were solved to be able to solve this question. So, we will be calculating the slip factor as well as the Mach number in this question that we are going to solve. Let us take a look at the problem statement for this question. A centrifugal compressor with a backward leaning blades develops a pressure ratio of 5 is to 1 with an isentropic efficiency of 83 percent. The compressor runs at 15000 rpm, induces are provided in the inlet of the compressor. So, that air enters at an absolute velocity of 120 meters per second. The inlet stagnation temperature is 250 Kelvin and the inlet air is given a pre whirl of 22 degrees to the axial direction at all radii. The mean diameter of the i of the impeller is 250 millimeters and the impeller tip diameter is 600 millimeters. Determine the slip factor and the relative Mach number at the impeller tip. So, you can immediately see that it has components of both the second question as well as third question. Second question we actually calculated the slip factor and in third question we calculated the absolute Mach number. Of course, in this case we are required to calculate the relative Mach number. Let us look at the velocity triangles first both at the inlet as well as the exit. This is the velocity triangle that you should get at the inlet and you have a schematic of the inducer and also a fixed inlet guide vane which gives a pre whirl to the inlet flow which is entering the inducer. So, these guide vanes put which are set at an angle of alpha 1 gives a pre whirl which is given in this case as 22 degrees causing the absolute velocity not to be axial. And therefore, the absolute velocity C 1 is not axial unlike some of the questions we have solved earlier on. And that causes the velocity triangle at the inlet to look like what is shown here. This is C 1 the relative velocity v 1 and the blade speed u 1. At the exit on the other hand these are backward leaning blades and so the velocity triangle for a backward leaning blade should look like this. Since, the blades are leaning like this the relative velocity leaves the blades in this direction with an angle of beta 2. And this is the absolute velocity C 2 and the blade speed u 2 at the tip of the impeller. The axial velocity component is shown here as C A and the rotational speed is omega which is given in this case as 15,000 r p. Let us first try to solve the inducer part of the question and then we will move towards solving the second part. Now, the inlet temperature is given as 300 Kelvin and therefore, based on the pressure ratio and isentropic relation we can calculate the exit stagnation temperature isentropic. And that is T 0 2 S which is equal to T 0 1 into pi c raise to gamma minus 1 by gamma. And this is 250 multiplied by 5 raise to 0.4 by 1.4 that is gamma minus 1 is 0.4 divided by 1.4 it is basically 395.95 Kelvin. Therefore, the stagnation temperature isentropic is 395.95 minus 300 that is 95.95 Kelvin. But the actual temperature rise is this divided by the efficiency. In this case efficiency is given as 0.83 and therefore, T 0 S actual basically would be equal to this divided by efficiency that is 95.95 divided by 0.83 that is 115.6 Kelvin. So, the specific work required would be C p times delta T 0 that is 1005 into 115.6 this is 116.186 kilo joules per kilogram. Now, since it is given that C 1 is 150 we can calculate C w 1 which is the world component of the absolute velocity that is C 1 times sin alpha 1 alpha 1 is 22. Let me go back to the velocity triangle here C 1 times sin alpha 1 would give us C w 1 which is this component. So, that comes out to be 56.2. Now, here what we are going to do is since we know the specific work done from the temperature rise we also know that specific work done is equal to the product of u 2 times C w 2 minus u 1 times C w 1. C w 1 we have calculated we can now calculate u 1 and u 2. Therefore, we get C w 2 and once we know C w 2 we can calculate the slip factor which is C w 2 by u 2. So, let us first calculate u 1 and u 2 u 1 is pi d mean of the i of the impeller into the rotational speed by 60 that is given as pi into 0.25 into 15000 by 60 this is 196.3 meters per second and u 2 is the blade speed at the tip of the impeller pi d t into n by 60 that is pi into 0.6 into 15000 by 60 471.24 meters per second. Now, specific work done as we have seen we calculated that already from the previous calculation specific work is C p delta t naught that is 116.186. Specific work is 116.186 into 10 raise to 3 is equal to u 2 which is 471.24 into C w 2 minus u 1 which is 196.3 into C w 156.2. So, from this we can calculate C w 2 that is 269.96 meters per second. Therefore, slip factor is the ratio of this C w 2 divided by u 2 that is 269.96 divided by 471.24 that is 0.573. You can see that the slip factor is a very low number here one would normally expect a relatively higher slip factor of the order of 0.7.8 or even higher than that. This is a very low slip factor and we have already discussed the disadvantages of having very low values of slip factor basically affecting the pressure rise of the centrifugal compressor for a given rotational speed. So, the next part of the question is to calculate the Mach number at the tip in the relative frame. So, the relative Mach number at the tip of the impeller for which we will refer to the velocity triangles once again. Let us look at just the exit velocity triangle. For a backward leaning blade we have seen that the velocity triangle would look like this. We have the absolute velocity C 2 and the relative velocity V 2 which is leaving the blades tangentially. So, this is how the exit velocity triangle would look like we need to calculate V 2 and also the temperature at the exit of the impeller to calculate the Mach number. So, from the velocity triangle we see that V 2 is equal to square root of C A square plus U 2 minus C w 2 the whole square. So, this C A at the inlet of course, we are assuming that the axial velocity does not really change as it passes through this impeller. This is equal to C A square that is C 1 cos alpha 1 the whole square minus well plus U 2 minus C w 2 the whole square. Since, all these parameters are known we substitute that we get the relative velocity 222.9 meters per second and then we need to also calculate the static temperature at the tip T 2 which is T 0 2 minus C 2 square by 2 C p. T 0 2 we can calculate because we know the efficiency and the pressure ratio. So, from there we calculate T 0 2 which is T 0 1 plus T 0 2 s minus T 0 1 by efficiency and that comes out to be 365.61 and we also need to calculate C 2 from the velocity triangle we see that C 2 is C 2 square is equal to the this component C w 2 square plus C A square. So, C w 2 we have already calculated that is 269.9 that square plus C A square which we can calculate from this C 1 cos alpha 1 that is 139.08 square. So, C 2 comes out to be 303.68 meters per second. Therefore, the static temperature is equal to 365.61 minus C 2 square by 2 C p therefore, T 2 comes out to be 319.73 Kelvin. Therefore, the relative Mach number is the relative velocity divided by square root of gamma R T 2 and this if you substitute we get a relative Mach number of 0.62. So, the relative Mach number at the impeller tip in this case is calculated as 0.62. So, this completes four problems that we have solved in today's class. We started off with a very simple problem which just involves solving the velocity triangle and calculating the velocities and also the angles involved. Second question was to calculate basically the slip. Third question involved calculating the Mach number in the absolute frame and the last question was a combination of the second and third calculating the slip factor as well as the Mach number at the tip. So, these were four questions that we have solved in today's lecture. I now have a few exercise problems which you can take up and solve based on our discussion today as well as what we have discussed during the lectures. Let us take a look at the first exercise problem. The first exercise problem states that the design mass flow rate of a centrifugal compressor is 7.5 kg per second with the inlet stagnation temperature of 300 Kelvin and pressure of 100 kilo Pascal. The compressor has straight radial blades at the outlet. The blade angle at the inducer inlet tip is 50 degrees and the inlet up to tip ratio is 0.5. The impeller is designed to have a relative Mach number of 0.9 at the inducer inlet tip. If the tip speed is 450 meters per second, determine part a the air density at the inducer inlet, part b inducer inlet diameter, part c the rotor rpm and part d the impeller outlet diameter. The answers to these four different parts of the questions are density should come out to be 0.988 kgs per meter cube, inducer inlet diameter is 0.258, rotor rpm is 17100 rpm and impeller outlet diameter is 0.502 meters. The second exercise question is a centrifugal compressor runs at 10000 rpm and delivers 600 meter cube per minute of air at a pressure ratio of 4 is to 1. Isentropic efficiency of the compressor is 0.82. The outlet radius of the impeller is twice the inner radius, axial velocity is 60 meters per second. If the ambient conditions are 1 bar and 293 Kelvin, determine part a the impeller diameter at inlet and outlet, the power input the and the impeller angles at the inlet. The answer to the questions are impeller diameter at the inlet should be 0.92 meters and outlet 0.461 and the power input is 2044 kilo watts, impeller and diffuser angles at the inlet are 13.9 degrees and 7.1 degrees. Third exercise problem is 30 kilograms of air per second is compressed in a centrifugal compressor at a rotational speed of 15000 rpm. Air enters the compressor axially and the compressor has a tip radius of 30 centimeters. Air leaves the tip with a relative velocity of 100 meters per second at an angle of 80 degrees. Assuming an inlet stagnation pressure and temperature of 1 bar and 300 Kelvin respectively, find part a the torque required to drive the compressor, the power required and the compressor delivery pressure. So, the torque in this case will be 4085 Newton meters, power required 6.417 megawatts and compressor delivery pressure is 6.531 bar. And the last problem is a centrifugal compressor has an impeller tip speed of 366 meters per second. Determine the absolute Mach number of the flow leaving the radial vanes of the impeller when the radial component of velocity at impeller exit is 30.5 meters per second and the slip factor is 0.9. Given that the flow area at the impeller exit is 0.1 meter square and the total to total efficiency is 90 percent determine the mass flow rate. So, in this case we have the absolute Mach number as 0.875 and mass flow rate as 5.61 kg per second. So, these are 4 exercise problems that you can solve based on what we have discussed in the last 3 lectures including today's. And I hope based on these discussions you will be able to solve these 4 exercise problems that we have for you today. And we will continue our discussion on some of these topics especially some of the radial flow machines. We will probably be taking up the radial flow turbines in the coming lectures. So, we will discuss more to do with the radial flow machines in some of the coming lectures in the next few lectures.