Question

A steam turbine inlet is at 1200 kPa, 500°C. The exit is at 200 kPa, 275°C. What is the isentropic efficiency?

Answer 1

Inlet: h_in = 3476.28 kJ/kg, s_in = 7.6758 kJ/kg K 

 Exit: h_ex = 3021.4 kJ/kg, s_ex = 7.8006 kJ/kg K

Ideal Exit: 200 kPa, s = s_in = 7.6758 kJ/kg K ⇒ h_s = 2954.7 kJ/kg 

w_ac = h_in - h_ex = 3476.28 – 3021.4 = 454.9 kJ/kg

w_s = h_in - h_s = 3476.28 – 2954.7 = 521.6 kJ/kg

η = w_ac /w_s = 454.9/521.6 = 0.872