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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 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. (Refer Slide Time: 17:32) So, if we add up all the three different terms and divide by 2, we get the specific work done as 225, 142 meter squared per second squared. And, what we see is that, the fractional contribution of each of these three terms. The first term which is U square is .375. V square is .158 and C square is .467, that is, you can see that there is a fair share for all these three 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 whole squares, you get 2 25,150 meter squared per second squared. 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 becausePDF Image | Turbomachinery Aerodyanmics
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