Turbine Isentropic Efficiency

The ideal turbine expands at constant entropy. It is the best a machine can possibly do between a given inlet state and a given exhaust pressure, it is a vertical line on the Mollier chart, and it is the yardstick everything else is measured against. Work the example through. Steam at 4 MPa and 400 °C has h = 3,214 kJ/kg and s = 6.769 kJ/(kg·K). Expand to 0.1 MPa holding that entropy: at 0.1 MPa the table gives $s_f = 1.3026$ and $s_{fg} = 6.0568$, so the ideal end point has quality (6.769 − 1.3026)/6.0568 = 0.9025, and its enthalpy is 417.5 + 0.9025 × 2,258.0 = 2,455 kJ/kg. The ideal drop is therefore 759 kJ/kg. If the machine really delivers 614 kJ/kg, its isentropic efficiency is 614/759 = 0.81.

Eighty-one percent is a normal answer. A modern multistage condensing turbine runs 80 to 90%, a well-matched back-pressure machine 70 to 80%, and a small single-stage unit driving a pump may only manage 50 to 65%. Friction, leakage past the blade tips, throttling at the governor valve and moisture drag all show up in that one number, and the moisture part follows the old Baumann rule of roughly one percent of efficiency lost per one percent of average exhaust wetness. If this page hands you a figure above 100%, treat it as information rather than a windfall: no expansion beats the isentropic one, because irreversibility only ever adds entropy and pushes the real end point to the RIGHT on the chart, which means to a higher enthalpy. An answer over 100% says the state points are wrong. The usual culprits are $h_{2s}$ evaluated at the wrong exhaust pressure, an entropy carried at too few figures, or an exhaust temperature measured downstream of a gland leak or a spray.