Steam Turbine Specific Work

Write the steady-flow energy equation across a turbine, drop the kinetic and potential terms because they are tiny, and drop the heat loss because a lagged casing is adiabatic to about one percent, and what is left is beautifully plain: the work out per kilogram is the enthalpy in minus the enthalpy out. Steam entering at 4 MPa and 400 °C carries 3,214 kJ/kg. If it leaves at 2,600 kJ/kg, the machine has taken 614 kJ out of every kilogram that went through, and no property other than those two enthalpies is involved.

Turbine people usually flip that into steam rate, which is the mass of steam needed per unit of work: 3,600 divided by 614 kJ/kg gives 5.86 kg per kilowatt-hour. It is the number a plant engineer carries in his head because it converts a load directly into a boiler duty. Two cautions. The exhaust enthalpy $h_2$ is rarely measurable, because a wet exhaust has no unique temperature at its pressure, so in practice $h_2$ is worked back from a measured power output or from an assumed isentropic efficiency rather than read off a thermometer. And this is specific work at the steam path, not at the coupling. Bearing friction, the governor, the oil pump and windage all come off before the shaft turns anything, and on a small machine those can be a few percent.