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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 diameters. So, I think we should 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 IFR 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 is U 2 square minus U 3 square. The second term was a function of the relative velocity and 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 three different terms, U 2 squares minus U 3 square plus W 2 square V 2 square minus V 3 square. And, the third term was the absolute velocity. So, let get these individual terms first, add them up and then we get the specific work done. (Refer Slide Time: 13:14) 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 revolutions per minute, the diameter is given as 23.76PDF Image | Turbomachinery Aerodyanmics
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