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A steam turbine inlet is at 1200 kPa and 500°C. The actual exit is at 300 kPa having an actual work of 407 kJ/kg. Find its second law efficiency?

A steam turbine inlet is at 1200 kPa and 500°C. The actual exit is at 300 kPa having an actual work of 407 kJ/kg. Find its second law efficiency? Correct Answer 0.98

To = 25°C = 298.15 K, hi = 3476.28 kJ/kg; si = 7.6758 kJ/kg K he = hi – w(ac) = 3476.28 – 407 = 3069.28 kJ/kg Te = 300°C; se = 7.7022 kJ/kg K wrev = (hi – Tosi) – (he – Tose) = (hi – he) + To(se – si) = (3476.28 – 3069.28) + 298.15(7.7022 – 7.6758) = 407 + 7.87 = 414.9 kJ/kg second law efficiency = w(ac)/w(rev) = 407 / 414.9 = 0.98.

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