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(Refer Slide Time: 25:04) So, the stagnation temperature drop, which is given by T 0 1 minus T 3 s is equal 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, rise 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 kilograms per second. C p, we are assuming for gasses as 1.148. Therefore, T 0 1 minus T 0 3 is 149.33 kelvin. So, if we simply take of ratio, this total-to-static efficiency is 149.33 kelvin 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 are 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 formula that we had derived in the last class. If you remember it was a very long formula, which was relating the last coefficient to the radius ratio and the static temperature ratios.PDF Image | Turbomachinery Aerodyanmics
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