Strip the steady-flow energy equation down to a turbine and almost everything cancels: no heat crosses the casing worth counting, and the velocity terms are small. What is left is the enthalpy drop. Read aloud: w equals h-one minus h-two — $w = h_1 - h_2$, where $h_1$ is the enthalpy at the stop valve in kJ/kg, $h_2$ is the enthalpy leaving the last stage in the same units, and $w$ is the specific work, the work taken from every single kilogram. Subscript 1 is always the inlet, 2 always the exhaust, everywhere in this chapter.

Then scale it up: $P = \dot m\,w$ — P equals m-dot w. $\dot m$ (the dot means per unit time) is the steam mass flow in kg/s, and $P$ is the shaft power at the coupling. Push the units through and the arithmetic explains itself: kJ/kg × kg/s leaves kJ/s, and a kilojoule per second is a kilowatt. That cancellation is the entire units stage of this lesson, done in your head.

Units guide, they never confess. If a rearrangement's units refuse to cancel into kW, the rearrangement is wrong and there is no appeal. Units that do work out prove nothing at all — the check runs one way only.