Hello and welcome to lecture number 17 of this lecture series on jet aircraft propulsion. In our journey so far, we have had quite some interesting revelations about the different components that constitute a gas turbine engine. And besides just the basic thermodynamics of these components, we have also had some discussions on how these components, different components are working. We of course, have not yet covered the rest of the components, we are on the axial compressors right now. Subsequently, we will of course, be taking up the other components one by one. And towards the end of the course, I am sure you would have a better understanding of how a gas turbine engine works and how is it that we can initiate a very preliminary design of these components. So, what we are going to discuss today is basically some solve problems on axial compressors, because we have been discussing about axial compressors in the last 3 or 4 lectures. And therefore, it is very much a good idea that we should be able, we should try and solve a few problems. And see how is it that we can apply the principles, the thermodynamic principles that we have studied in the last few lectures on trying to solve a few problems. So, that we get some idea about what the numbers look like for an axial compressor. So, what I have for you today are a few solve problems, I basically sorted out 3 problems which I shall solve for you. And then I have also have listed a few problems towards the end of the lecture which are basically exercise problems for you to solve based on our discussion today as well as what we have been discussing in the last few lectures. So, in today's tutorial we shall be discussing about solve problems on axial compressors. So, today's lecture is basically a tutorial with both solved as well as unsolved problem towards the end of the lecture. Let us take a look at the first problem that we have for today's lecture. So, this problem is on again an axial compressor which states that air at 1 bar and 288 Kelvin enters an axial flow compressor with an axial velocity of 150 meters per second. There are no inlet guide wings, the rotor stage has active diameter of 60 centimeter and a hub diameter of 50 centimeters and rotates at 100 revolutions per second. The air enters the rotor and leaves the stator in an axial direction with no change in velocity or radius. The air is turned through 30.2 degrees as it passes through the rotor. Assume a stage pressure ratio of 1.2 and overall pressure ratio of 1.6. Find part A the mass flow rate of air, part B the power required to drive the compressor, part C the degree of reaction at the mean diameter, part D the number of compressor stage is required if the isentropic efficiency is 0.85. So, in this problem it is basically stated that we have some inlet conditions, we have an axial velocity which has been specified and it is clearly mentioned that there are no inlet guide wings. So, what does this mean? It basically means that the axial velocity and the absolute velocity at the inlet are in the same direction which will be clear when we take a look at the velocity triangles as well. And besides this we have some geometric details given and how much is the air deflected through as it passes through the rotor. So, the deflection angle is given and besides this we have been given the pressure ratios. So, we are required to find the number of parameters, the number of stages required, power required and so on. So, I think the first step towards solving any such problem is to draw the velocity triangles. And I think if you get the velocity triangles right then that solves half of the problem. And that is very important in solving such problems because from the velocity triangle it will be very clear that what are the different components that which have been specified in the problem and what are the other components that you need to find to be able to fully solve this problem. So, you should definitely begin with drawing the velocity triangle and that is very important. And based on of course, the information provided for example, in this question it is given that there are no inlet guide wings. So, how will the velocity triangle look like? So, we will take a look at what the velocity triangle looks like at the inlet of the rotor that is at the leading edge. And at the trailing edge again we will have to figure out what the velocity triangle should look like. And some of the components in the velocity triangle would be known. The blade speed is given as 100 revolutions per second and the diameter at which we have to calculate these properties are also known. So, based on that we can calculate the blade speed and the inlet axial velocity is given and so on. And also the turning angle through the rotor is specified. So, based on the information that we have let us now construct the velocity triangle. So, velocity triangle for this particular problem would look something like this. Now, let me explain why the velocity triangles have been constructed like this.  So, we begin with constructing a blade compressor blade it is typically shown here. A low speed compressor blade is what is shown here. And the blade speed which is basically u has been specified in this direction. So, this is u the blade speed because it is a compressor blade we know the direction of u. In this question we have been we have been given that the axial velocity is 150 meters per second at the inlet and there are no inlet guide vanes. So, if there are no inlet guide vanes what it means is that whatever velocity is coming in it enters the blade axially. That is the inlet velocity the inlet absolute velocity that is c 1 will be equal to c a and it will be in the same direction. This is because there are no inlet guide vanes. And velocity v 1 will always be in a direction which is tangential to the blade leading edge. So, v 1 is tangential there and u is known. And so we know that vectorially u plus v 1 is equal to c and that is how you get the inlet velocity triangle. And the angle which v 1 makes with the axial direction is given by beta 1 which is the blade inlet angle. So, this is how the velocity triangle is constructed at the inlet. And what happens at the exit or at the trailing edge at the trailing edge again we know that v 2 that is the relative velocity should leave the blade tangentially. So, this is v 2 the direction of the relative velocity. And we also know the blade speed blade speed is u which is known the direction is known magnitude is also known. And that completes the exit velocity triangle because u plus v is equal to c. So, u 2 u plus v 2 is equal to c 2 and therefore, that completes this velocity triangle. And beta 2 is the blade angle at the exit that is at the trailing edge the angle made by the relative velocity with the axial direction. So, that is given as beta 2. So, these are the different components of the velocity triangle. And once we get the velocity triangle we would also be able to identify what are the parameters which are known. In this case for example, we know the inlet axial velocity that is specified we know the blade speed. And so, based on that with we know a certain parameters of this velocity triangle. And we should now start to try and solve the problem and try to find out the other components of the velocity triangle. And subsequently of course, we will need to also find out the other parameters like the pressure ratio and so on and efficiency. So, let us start to solve this problem. Now, how do you find for the first parameter to be calculated is the blade speed at the mean diameter. We have been given the tip diameter and the root diameter. So, based on that we can find out the blade speed blade speed is u which is pi into mean diameter divided by 2 into n which is the speed given in revolutions per second here. So, here we have pi into d t plus d h which is the tip diameter plus the hub diameter divided by 2 and multiplied by n. So, we get pi times the tip and root diameters divided by 2 into 100 which is the blade the revolutions per second. So, if you calculate that we should be able to get the blade speed at the mean diameter as 172.76 meters per second. So, we have now found out the blade speed we also know the axial velocity. So, at the inlet velocity triangle we now know two parameters we know the axial velocity which is 150 meters per second and u we have just now calculated as 172 meters per second. So, based on this we can find out the angle beta 1, beta 1 tan beta 1 is basically u by C A and therefore, beta 1 is tan inverse u by C A. So, if you were to use that find out beta 1 as tan inverse u by C A we get that as 49.2 degrees and beta 2 is the exit blade angle blade angle at the exit of the rotor and in this question it is given that the blade turns the flow by 30.2 degrees. And therefore, since beta 1 is 49.2 and the turning angle which is or deflection angle which is delta beta delta beta is 30 degrees 30.2 degrees. Therefore, we have beta 2 which is 49.2 minus 30.2 which is 19 degrees. So, we have now calculated the blade angle at the inlet and exit of the rotor. So, we now have beta 1 and beta 2. So, now if we know beta 2 we should also be able to calculate some of the other parameters. For example, tan alpha 2 which is the angle at the exit of the absolute velocity that is u minus tan alpha 2 is u minus C A into tan beta 2 divided by C A that is this is u tan alpha 2 is this component alpha 2 that is u minus C A into tan beta 2 which is this component divided by C A. Because it is given that there is no change in axial velocity as it passes through the rotor. So, from there we can calculate tan alpha 2 and therefore, alpha 2 which comes out to be 38.92. Remember that at the inlet alpha 1 and beta 1 are the same because there is no angle made by the absolute velocity with the axial direction because C A is equal to C 1. So, beta 1 and alpha 1 are the same which is 49.2. So, alpha 2 comes out to be 38.92 degrees. So, we have calculated all the angles we have now calculated beta 1 we now we have calculated beta 2 we have calculated alpha 2. And so, we have calculated all the angles required and so that should be specifying the geometry in terms of the angles. So, let us now calculate the other parameters that are required. And so, the second part of the question was to find out let us look at what is the second part question requires that we have we required to find out the mass flow rate of air. Then we find out the power required at degree of reaction and number of stages. So, let us calculate or determine the mass flow rate. How do we calculate mass flow rate? Mass flow rate is given by the area that is the annular pi by 4 into d t square minus d h square into velocity in into the density that is at the exit. So, we can either find out the density at the exit or at the inlet correspondingly use the actual velocities. So, we have here then the temperature T 1 which is equal to T 0 1 minus C a square by 2 C p. So, T 0 1 is given in this question it is already been specified C a is already known. So, we can find out T 1 that comes out to be 276.8 Kelvin. And the stagnation temperature we can find by the isentropic relation assuming that because there is no efficiency specified here we can assume that the process is isentropic. So, T 0 2 that is stagnation temperature at the exit of the rotor is equal to T 0 1 into P 0 2 by P 0 1 raised to gamma minus 1 by gamma. And this is the compressor pressure ratio that is already given as 6. So, from there we can find out the temperature at the exit of the rotor and that comes out to be 303.41 Kelvin. So, once we find out T 0 2 we can find out T 2 that is the static temperature that is T 0 1 T 0 2 minus C 2 square by 2 C p. And so how do we find out C 2 square C 2 is not known and C 2 is basically equal to C a by cos alpha 2 because let us go back to the velocity triangle C 2 is this component C 2 is equal to C a that is this divided by cos alpha 2. And cos alpha 2 we have already found out that is 38.92. So, C 2 we can find out as 150 divided by cos 38.92 that is 192.79 meters per second. Therefore, we find out T 2 that is static temperature that is 303.41 minus C 2 square by 2 C p that is 192.79 square by 2 into 1005 that is 2010. So, this comes out to be 284.91 Kelvin. Similarly, we find out P 2 which comes out to be 1.2 bar because the stage pressure ratio is given and so from there we can find out P 2. And so once we find out pressure and temperature we can find out now the density is equal to P by RT. So, P we have converted that is in bars we converted that to Pascals divided by R that is gas constant times the temperature. So, the density at the exit of the rotor comes out to be 1.507 kilograms per meter cube. So, from this equation that we have written mass flow rate is basically pi by 4 into dt square minus d h square the tip and hub diameters are already specified C 2 we have determined and rho 2 also we have calculated. So, if you substitute all those values the mass flow rate comes out to be 19.53 kilograms per second. So, that solves the first part of the problem. The second part of the problem was to find out the power required to drive this compressor and how do we find out power required? Power required is u into delta C theta multiplied by mass flow rate. And delta C theta is or C w is basically the difference between these two components that is this is C w 2 and this is C w 1. So, from there basically what happens is we have it is basically the delta C w is equal to C a times tan beta 1 minus tan beta 2 and u is already been calculated 172.76 C a is known 150 meters per second mass flow rate is 19.53 and tan beta 1 and tan beta 2 are also known. Therefore, the power required can be calculated by substituting all these values and simplifying. So, we get the power required as 412 kilowatts. So, so far we have calculated two parts of the question one was to find out the mass flow rate which is basically the annular area times the velocity multiplied by the density. And you could either find that out at the inlet or exit. So, that is up to you whichever way you can find out. But obviously, the mass flow rate will come out to be the same. The second part of the question was to find out the power required. And power required is basically equal to mass flow rate multiplied by the blade speed that is u and delta C w. In this case it comes out to be C a times tan beta 1 minus tan beta 2. So, from this we find out the power required as well. So, the next part of the question is to find out the degree of reaction for this particular rotor. Now, if you recall our discussion on degree of reaction a few lectures ago, degree of reaction is basically the ratio of the work done by the rotor for the overall compression as part of the stage. So, that is static enthalpy rise in the rotor divided by the stagnation enthalpy rise in the stage. So, that defines the degree of reaction. And we have simplified that basic definition to arrive at an expression which can be expressed in terms of known parameters like the blade speed or the angles and so on. So, if we look at what is degree of reaction degree of reaction is basically r r x as we have defined is equal to 1 minus C a by 2 u into tan beta 1 plus tan beta 2. So, we have already derived this equation in our earlier lecture. So, if you use that expression substitute all the values because all these parameters are known. So, if you were to substitute those values we have 1 minus C a which is 150 divided by u or 2 u that is 2 into 172.76 multiplied by tan beta 1 that is tan 49.2 plus tan beta 2 that is tan 19. So, this comes out to be 1 minus 0.65 that is 0.35. So, this is the degree of reaction for this particular rotor. Now, the next part of the question is to find out the number of stages required by this compressor to generate this particular pressure ratio of 6. So, for finding out the number of stages what we are going to do is which is something I think you would also be doing later on is that the overall pressure ratio is known. And so we can calculate the overall temperature rise in this particular compressor that is delta T naught overall because the pressure ratio is known. We can also find out the per stage temperature rise because we have already defined per stage pressure rise or temperature rise in terms of the blade speed and the axial velocity and the geometric angles. So, total temperature rise across the compressor divided by per stage temperature rise will give us will tell us how many number of stages are required for generating this pressure ratio. So, first thing we will calculate is delta T naught for one stage. So, we have already looked at this equation we have discussed this equation earlier how we have arrived at this particular expression. So, delta T naught for the stage is u times C a divided by C p into tan beta 1 minus tan beta 2. Basically, this comes from the fact that we equate delta H naught to u times delta C w. So, if we equate the two we get the same expression that is C p into delta T naught should be equal to u times C a into this. So, let us substitute for all these values when we substitute them we know u that is 172.76 C a is known as 150 C p is already specified tan beta 1 and tan beta 2 are known. So, we get delta T naught for the stage as 20.99 Kelvin. So, this is the stagnation temperature rise occurring in one stage of this axial compressor and delta T naught overall we can find out we basically the overall temperature rise is the inlet temperature that is T 0 1 divided by the efficiency because it is given as 0.85 multiplied by the pressure ratio overall pressure ratio raise to gamma minus 1 by gamma minus 1. So, that is 288 by efficiency isentropic efficiency is 0.85 multiplied by 6 that is the overall pressure ratio raise to gamma minus 1 by gamma that is 1.4 minus 1 by 1.4 that is 0.286 minus 1. So, delta T naught overall is 226.5 Kelvin. Therefore, the number of stages required for generating a pressure ratio of 6 would be overall temperature ratio or overall temperature rise delta T naught overall divided by delta T naught for the stage. So, that is 226.5 divided by 20.99 that is 10.79 which is equal to 11. So, 10.799 we have rounded it off to the next higher digit that is 11 because if you use only 10. Obviously, you cannot achieve the required pressure ratio of 6. So, that this is a normal practice that if you arrive at a fractional number there we round it off to the next higher fraction number. So, that we are able to achieve the desired pressure ratio. So, the number of stages we have calculated here by calculating the overall temperature ratio or temperature rise divide that by the temperature rise required for or generated by one stage. So, that gives us the number of stages required for generating this pressure ratio of 6 in this case. So, in this particular problem that we have solved we have basically been discussing about the axial compressor and how we can go about calculating the various parameters required in this particular case. For example, in the first problem that we have solved we have been specified some of the angles we have given the inlet flow conditions based on that we have calculated the mass flow rate, the power required and the degree of reaction and of course, the number of stages required. So, we will now take a look at the next problem that we have which is also pertaining to an axial compressor. Let us take a look at what the problem statement is. Now, this problem which has been specified here is on an axial compressor which is to be designed to generate a pressure ratio of 4 with an overall isentropic efficiency of 0.85. The inlet and outlet blade angles of the rotor blades are 45 degree and 10 degree respectively and the compressor stage has a degree of reaction of 50 percent. If the blade speed is 220 meters per second and the work done factor is 0.86, find the number of stages required and you also need to find out is it likely that the compressor will suffer from shock losses. The ambient air static temperature is 290 Kelvin and the air enters the compressor through guide ways. So, this question that we have pertains to a rotor which is having a degree of reaction of 50 percent. Now, what is the property of a rotor blade which has a degree of reaction of 50 percent? We have discussed that if degree of reaction is 50 percent the velocity triangles will be symmetrical or mirror images that is alpha 1 will be equal to beta 2, alpha 2 will be equal to beta 1 and we have been given 2 of the angles here which means all the angles of the velocity triangle are already known and so based on this we have to now find out the number of stages required and we also need to find out whether this particular rotor will be suffering from any shock losses. So, as we discussed in the last question the first step towards solving such a problem is to get the velocity triangles. In this case the velocity triangles are relatively easy because it is given that it is a 50 percent reaction rotor that is velocity triangles are exactly symmetrical. So, with that in mind we have the velocity triangles which are exactly symmetrical here and therefore, this means that since the velocity triangles are symmetrical we will have alpha 1 which is equal to beta 2 and beta 1 is equal to alpha 2 and also that will mean that some of these velocity components are also equal c 1 will be equal to v 2 c 2 is equal to v 1 and so on. And so, we have the inlet and outlet blade angles of the rotor that is beta 1 is 45 and beta 2 is 10 which means the corresponding angles alpha 2 will be 45 and alpha 1 will be 10. The blade speed is already given as 220 meters per second and there is another factor that has come up here that is known as work done factor. I will discuss that shortly what is meant by work done factor. So, axial velocity we have been given blade speed. So, how do we find out axial velocity? So, axial velocity we can find out from the velocity triangles. Now, in the inlet velocity triangles we basically let us look at the axial velocity at the inlet or exit both are same. So, c a is this component. So, tan alpha 1 or tan beta 1 for that matter will be this component and the rest of it will be from what is remaining of u. So, if we combine both the equations at the inlet velocity triangle and the exit velocity triangle then we get the axial velocity as u divided by tan alpha 1 tan beta 1 plus tan beta 2. So, let us see how that has come up. So, u is basically equal to c a tan beta 1 plus c a tan beta 2. So, if you look at u that is c a is this component and tan beta 2 or tan beta 1 is this component and tan beta 2 is this component. So, we are adding up these two and that is how we get c a is basically u divided by tan beta 1 plus tan beta 2 and. So, since tan beta 1 or beta 1 and beta 2 are known we can calculate the axial velocity that is 187 meters per second. Now, let us also calculate the absolute velocity at the inlet absolute velocity at the inlet is c a divided by cos alpha 1. Let us see how that comes c 1 is c a divided by cos alpha 1 cos alpha 1 is or alpha 1 is given as 19 degrees. So, if you substitute that we get c a or c 1 as 190 meters per second. Now, how do we calculate the per stage temperature rise because we have to now find out the number of stages required for this compressor. So, per stage temperature rise we have earlier defined as u divided by c u into c a divided by c p multiplied by tan beta 1 minus tan beta 2. Now, in this case I have included another factor here which is given by lambda this is basically known as the work done factor. So, work done factor is like an efficiency parameter which has been added basically because if we consider a multi stage compressor as we proceed from the inlet of the compressor to the exit of the compressor. I was discussing about annulus loss that is the growth of boundary layer from the inlet to the exit that is there is a boundary layer growth or development taking place from the inlet or the exit. And that means that the effective area of the flow that is available which is free of the boundary layer will continuously reduce that is the blockage increases as we move from the inlet of the compressor to the exit there is an increase in blockage as we go from inlet to exit. So, because of this increase in blockage we are accounting for that by using what is known as a work done factor that is from as we move from inlet to the exit of the compressor there is an increase in the boundary layer thickness that means that the effective area that the flow sees will be reduced as we move from inlet to exit because there is a growth of boundary layer that is occurring which means that it is basically appearing as a blockage for the potential flow. So, one of the ways of accounting for that is to use what is known as a work done factor that is because of this effective blockage there is per stage temperature rise is reduced. And therefore, the per stage pressure ratio is also decreased as a result of this blockage. So, work done factor is basically taking into account the fact that there is a reduction in the area or there is an increase in blockage as the stage proceeds from the inlet to the exit. So, per stage temperature rise the expression for per stage temperature rise will also have this work done factor unless otherwise specified work done factor is assumed to be 0 to be 1. So, if you look at the per stage temperature rise now we have this work done factor lambda multiplied by u C a by C p into tan beta 1 minus tan beta 2. So, if you substitute all the values work done factor is given u is known C a has already been calculated beta 1 and beta 2 are known and C p is also known. So, if you substitute all these values here we have the per stage temperature rise which is 29 Kelvin. So, let us now calculate the actual temperature rise across the compressor. So, here we have the total temperature at the compressor inlet we have T 0 2 that is inlet compressor temperature total temperature which is static plus C 1 square by 2 C p. So, that is 331.8 Kelvin because we know the inlet static temperature and we also know the absolute velocity at the inlet. So, we calculated the total temperature that is 331.8 Kelvin. Now, overall pressure ratio is given as 4 which means that the isentropic temperature at the compressor exit can be found out from the isentropic relations that is T 0 3 s is equal to T 0 2 into pi C which is the compressor pressure ratio overall pressure ratio raise to gamma minus 1 by gamma. So, that is 493.9 Kelvin. So, that is the temperature the total temperature at the exit of the compressor if the compression process was to be isentropic, but we know that it is not isentropic because there is an efficiency specified. Therefore, we can find out the actual total temperature at the compressor exit which is T 0 3 that is T 0 2 inlet temperature plus T 0 3 s minus T 0 2 divided by efficiency. So, that comes out to be 522.5 Kelvin. So, this is the actual temperature at the compressor exit. So, we can now find out the total temperature rise across the compressor which is the difference between the total temperature at the exit and the total temperature at the inlet of the compressor. So, this is 190.74 Kelvin. So, across the compressor there is an increase in total temperature of the order of 190.74 Kelvin. So, we now have the overall temperature rise across the compressor we also have the per stage temperature rise which we had earlier calculated. And therefore, number of stages required would be the ratio of the overall temperature rise to the per stage temperature rise. So, that is 190.74 divided by 29 that is 6.6 and as we have been as we have discussed earlier this will be rounded off to the next integer. So, we get number of stages as 7. So, in this case we have the number of stages that are required to generate pressure ratio of 4 as specified in this problem as 7. So, 7 number of stages are required to generate an overall pressure ratio of 4. So, that was part 1 of the question to find out how many stages are required for generating this particular pressure ratio given these parameters and the velocity triangles and if the degree of reaction was 50 percent. Second part of the question is to find out if this particular rotor is likely to suffer from any shock losses. So, in this case what we are going to do is that we will basically find out the relative Mach number and see that if the relative Mach number has is greater than 1 point or is greater than unity that means that the flow at the exit is likely to be supersonic which means that there is a likelihood that there would be shocks present and therefore, shock losses. So, let us go back and look at the velocity triangle once again. So, in this velocity triangle we are going to calculate the Mach number based on the relative velocity. So, that is known as the relative Mach number and so Mach number based on this would be V 1 divided by square root of gamma RT 1 and if that exceeds 1 then we can infer that the flow is supersonic and there is a likelihood of shock losses there. So, we will calculate the relative Mach number based on the relative velocity. So, relative Mach number is basically the ratio of the relative velocity to the speed of sound and that is square root of gamma RT. And how do we calculate V 1? V 1 we can calculate because the axial velocity is known the angles are also known from the velocity triangle. And therefore, V 1 is the axial velocity divided by cos beta 1 and that is 264.5 meters per second. So, we get the relative velocity at the inlet as 264.5 meters per second. Therefore, the relative Mach number is V 1 divided by square root of gamma RT. So, that is 0.77. So, what we see here is that the relative Mach number is less than unity and so that is no chance that this particular compressor at least at this particular diameter which has been given is likely to suffer from any shock losses. So, it is a normal practice to find out the relative Mach number at the tip of the rotor. In this case, we do not know really that the data given is at least at the tip. But if it is indeed at the tip then the relative Mach number which is the tip relative Mach number that would be the parameter which would decide whether this compressor rotor would suffer from any shock losses or not because if the tip Mach number exceeds 1 then there are chances that the rotor might suffer from shock losses. So, this problem that we have just now solved was on a compressor which was working on a 50 percent degree of reaction stage. Now, we have the third problem for today that is again on an axial compressor and we have some of the parameters at the inlet which has been specified that is conditions at the entry of an axial compressor stage or P 1 is 1 bar and T 1 is 314 Kelvin. The angles at the as specified for this particular rotor are beta 1 is 51 degrees, beta 2 is 9 degrees, alpha 1 and alpha 3 is equal to 7 degrees. The mean diameter and the peripheral speed are 50 centimeters and 100 meters per second respectively. Given that the work done factor is 0.95, the stage efficiency is 0.88 and the mechanical efficiency is 0.92, the mass flow rate is 25 kgs per second. Determine part A the air angle at the stator entry that is alpha 2, part B the blade height at the entry and the hub to tip diameter ratio, part C the stage loading coefficient and part D the power required to drive the stage. So, in this question all the angles at the that is the blade angles are specified beta 1 and beta 2 as 51 and 9 degrees. We also have one of the inlet angles that is the at alpha 1 that is 7 degrees and that has been given as equal to alpha 3. So, one part of the question is to find out the angle that is alpha 2 at the stator entry and we need to find out the power required then we need to find out work the stage loading coefficient and so on. And we have been specified the mean diameter blade speed and the mass flow rate besides the stage efficiency and the mechanical efficiency. So, based on the data that has been given we need to find out the angle alpha 2 then the power required and of course, the stage loading coefficient. So, let us see how we can solve this problem. So, part A is to find out the air angle at the stator entry which is alpha 2. So, from the velocity triangle that we have seen earlier. So, it is the same velocity triangle that we have discussed in problem number 2. So, I have not repeated it here. So, we can infer from the velocity triangle that u by C A that is blade speed divided by the axial velocity is equal to tan alpha 1 plus tan beta 1 and alpha 1 and beta 1 are already specified as 7 and 51 degrees u is given as 100 meters per second and therefore, from this we can find out the axial velocity that is C A. So, C A comes out to be 73.65 meters per second. So, the same velocity triangle also tells us that tan alpha 2 plus tan beta 2 is also equal to u by C A tan beta 2 is already specified that is 9 degrees u is known and C A we have just now calculated. So, if you substitute for that here we get tan alpha 2 plus tan 9 is equal to 100 by 73.65 therefore, alpha 2 is 50.18 degrees. So, that is the angle at the stator entry. So, we have calculated alpha 2 which is basically the first part of the question to find out the angle at the stator entry. Now, second part of the question is to find out the blade height at the entry and also the hub to tip diameter ratio. Now, to find out the blade height we have been we already know the mass flow rate and so mass flow rate is given as 25 kg per second axial velocity is known the inlet pressure and temperature are known therefore, we can find out the density mean diameter is known and therefore, rho times C A into the annular area that is pi into D into H is mass flow rate. So, if you substitute for mass flow rate the density axial velocity and the mean diameter we can calculate the blade height which is 0.19 meters and so we now have the mean diameter and the blade height and therefore, we can find out the tip diameter and the hub diameter and from there that is D T that is the tip diameter comes out to be 69 centimeters because mean diameter is given as 50 and the hub diameter is 31 centimeter which is again D T minus H. Therefore, the hub to tip diameter ratio is D H by D T that is diameter at the hub divided by diameter at the tip that is 0.449. So, that is basically tells us what is the annular area from the hub to tip diameter ratio. So, third part of the question is to find out the blade loading coefficient which is the work done divided by u square and we can find out work done based on our equation we had seen earlier that is u times delta C w multiplied by work done factor that is lambda. So, work done per stage is lambda times C A into u into tan beta 1 minus tan beta 2 which is basically u into delta C w. So, work done factor in this question is given as 0.95 C A we have calculated as 100 meters per second u is specified as 73.65 and tan beta 1 is tan 51 degrees and tan beta 2 is tan 9 degrees and so if you substitute these values we get the work done per stage as 7534.8 joules per kilogram. And therefore, the loading coefficient is work done per stage divided by u square and that is 7534.8 divided by 100 square. So, 0.7535 is the loading coefficient for this particular rotor which has been specified. So, loading coefficient is work done divided by u square and therefore, we have 7534.8 divided by 100 square that is 75 0.7535. And the last part of the question that is part d is power required to drive this compressor. So, power required is mass flow rate into the work done per stage divided by the mechanical efficiency. So, we have mass flow rate which is specified as 25 kgs per second work done we have already calculated 7534.8 joules per kilogram. And this divided by mechanical efficiency would give us the power required. And we are dividing by mechanical efficiency because mechanical efficiency represents the loss of power which has occurred while the power was being transferred to the compressor. That means, the compressor will require so much power extra to overcome the mechanical losses. And therefore, mechanical efficiency indicates losses in transmission from the turbine to the compressor. So, which is why the power required has been divided by mechanical efficiency. So, this is 204.75 kilowatts. So, this is the power required to drive this compressor. So, we have now solved 3 different problems pertaining to axial compressors. And in all the cases we have seen the basic problem solving should begin with the velocity triangle. And once the velocity triangle is specified and we mark the different components which are known and specified in the problem that makes the problem solving much more easier. And chances of making mistakes are less likely. And so, we have in all the 3 problems we have seen that given a certain geometric parameters and angles pertaining to a stage of the compressor. We can make use of the velocity triangles to solve the velocity triangle as well as to calculate and determine the other parameters like work done in fact work done per stage or power required or the other dimensions like the hub to tip diameter ratio and so on. So, there are 3 problems that we have solved in today's lecture so far. And I have a few exercise problems that problems for you which you can attempt to solve based on what we have discussed in today's lecture as well as our discussion during the earlier lectures. So, let us take a look at the first exercise problem that I have for you. So, the first exercise problem in for today is an axial compressor with 50 percent reaction design has blades with an inlet and outlet angles of 45 degrees and 10 degrees respectively. The compressor is to produce a pressure ratio of 6 is to 1 with an isentropic efficiency of 0.85 when the inlet static temperature is 37 degree Celsius. The blade speed and the axial velocity are constant throughout the compressor assuming a value of 200 meters per second for the blade speed find the number of stages required if the work done factor is part A that is unity and part B that is 0.87. So, in this question we have a rotor which has a degree of reaction of 50 percent which means that the velocity triangles are symmetric and mirror images 2 of the angles have been specified. So, all the angles of the velocity triangle are known and you have been given the blade speed and overall pressure ratio. And based on this we need to find out the number of stages required in 2 cases that is if the work done factor was unity. And second case was if work done factor was 0.87. So, answer to this question is if it if the work done factor is unity the number of stages required is 8 and if you take a work done factor of 0.87 the work done factor comes out to be 9. And so, you can see that there is an increase in the number of stages as the work as we start assuming values of work done factor because that reduces the per stage temperature rise which means that the number of stages required for generating the same pressure ratio which is given as 6 is to 1 in this case would be higher as a work done factor comes into play. The second exercise problem is stated as follows. Air at 1 bar and 288 Kelvin enters an axial flow compressor with an axial velocity of 150 meters per second and there are no inlet guide vanes specified the rotor has a tip diameter of 60 centimeter and hub diameter of 50 centimeter and rotated rotates at 100 revolutions per second. The air enters and the rotor and leaves the stator with no change in velocity or radius the air is turned through 30 degrees as it passes through the rotor. Determine part A the blade angles part B the mass flow rate part C the power required and part D the degree of reaction. So, this is a problem which is very similar to the problem that we solved in the first problem that we solved today. So, in this case the blade angles come out to be 49 degrees and 19 degrees mass flow rate as 14.38 kilograms per second part C as 30 300 that is power required as 300.7 kilowatts and degree of reaction as 0.65. Problem number three is axial flow compressor stage has the following data degree of reaction is 50 percent. The mean blade diameter is 36 centimeter rotational speed is 18000 rpm blade height at entry is 6 centimeters. Air angles at the rotor and stator exit are 25 degrees, axial velocity is 180 meters per second, work done factor is 0.88 stage efficiency is 0.85, mechanical efficiency is 96.7 percent. Determine part A the angle air angles at the rotor and stator entry part B mass flow rate part C power required part D the stage loading coefficient part E pressure ratio developed by the stage and part C part F that is relative Mach number at the rotor entry. So, the answer to this question that part A the angles are 54.82 degrees and 25 degrees the mass flow rate is 14.37 kilograms per second the power required is 51.2 kilo joules per kilogram and this stage loading is 0.44 pressure ratio developed is 1.6 and Mach number relative Mach number at rotor entry is 0.9. Question number four exercise problem number four is the 50 percent reaction axial flow compressor has inlet and outlet blade angles of 45 degrees and 12 degrees respectively. The blade speed at the tip of the rotor is 320 meters per second if the inlet total temperature is 300 Kelvin determine the tip relative Mach number. So, in this case the tip relative Mach number is 1.146 and the last problem for you to solve is a 10 stage axial flow compressor develops an overall pressure ratio of 8 with an isentropic efficiency of 0.85. The absolute velocity component of air enters the rotor at an angle of 27 degrees to the axial direction the axial component of velocity is constant throughout the compressor and is equal to 150 meters per second. If the ambient air conditions are 15 degrees Celsius and 1 bar determine the angle which the relative component of velocity makes with the axial direction at the exit of the rotor. So, this case the angle comes out that is beta 2 is 14 degrees. So, in today's lecture. So, there are five exercise problems for you to solve and the key to solving all these problem is to get the velocity triangles right because angles are specified in many of them and. So, you should begin with trying to get the velocity triangles for this these problems and then proceed towards solving these problems. So, I hope you would be able to solve these problems based on what we have discussed in today's lecture as well as what we have discussed in the last few lectures on theory and working of the axial flow compressors. So, that brings us to the end of today's lecture which is which was basically a tutorial session. In the next lecture we shall begin the next chapter and we shall be discussing about centrifugal compressors. So, we will begin with the discussion on elements of centrifugal compressors in the next lectures and we will take up centrifugal compressors in the next couple of lectures as well. So, that brings us to the end of today's tutorial session.


I'm Vicky DeNoya. I'm a nurse practitioner. I've been a nurse practitioner for 16 years and a registered nurse much longer. My subspecialty is osteoporosis. It's my passion. I have been doing this for 16 years, screening, preventing, evaluating older adults for their risk for fracture and treating and managing that. Osteoporosis is a disease of the bone. It's the most common bone disease. The bones lose their density and their strength, which increases a person's risk for fracture. Currently in this country, more than 54 million people have low bone mass and osteoporosis. However, only 10 million have been diagnosed or evaluated. This is pretty close to what normal bone tissue looks like. You can see it's pretty dense. This is the bone tissue. This is the outer covering of the bone. This model is the beginning of osteoporosis. On the DEXA test, this is a minus 2.6 which is the beginning of osteoporosis. I help patients manage their health by making sure that if they have healthy bones now, we want to make sure they continue to have healthy bones as they age. We want to make sure they are taking enough calcium every day, ideally in their diet. And if they're not getting it in their diet, they may need a supplement.  For men, it's 1,000 milligrams a day. For women, it's between 1,200 and 1,500 milligrams a day. If you're smoking, you really need to stop smoking. Tobacco increases your risk for osteoporosis. Make sure that you have a vitamin D tested periodically. And as you get older, make sure your height is measured. Usually people get their height measured in high school or college and that's it. They never know that they're losing height until, as patients say to me, my pants are getting longer. In January 2017, we opened the Center for Bone Health. At the Bone Health Center, the services we offer here, DEXA screening because the DEXA is the gold standard for screening osteoporosis. It is a full person evaluation. We order lab testing and if there's any indication that they might have a secondary cause of osteoporosis, then we refer either to a rheumatologist or an endocrinologist for further testing and follow up.