Steady-Flow Energy Balance: Nozzles, Turbines, Throttles

A control volume lets mass cross its boundary. For steady flow (nothing changes with time), the energy balance is the steady-flow energy equation (SFEE):

Q̇ − Ẇ = ṁ·[ (h2 − h1) + (V2² − V1²)/2 + g·(z2 − z1) ]

Dots are rates (kW), V is fluid velocity (m/s), and the (V²)/2 term is kinetic energy per unit mass (J/kg — watch the units).

Simplifying by device

Device	Typical simplification
Nozzle	Ẇ = 0, adiabatic, inlet velocity often neglected
Turbine	Ẇ is the output; Δke, Δpe usually small
Throttle (valve)	Q̇ = Ẇ = 0, Δke ≈ 0 ⇒ h2 = h1
Heat exchanger	Ẇ = 0; streams exchange enthalpy
Pump/compressor	Ẇ is the input; Δke, Δpe small

Worked example: nozzle exit velocity

Steam enters a nozzle at h1 = 3000 kJ/kg, V1 ≈ 0, and exits at h2 = 2700 kJ/kg. Adiabatic, no work. Find V2.

With Q̇ = Ẇ = 0 and V1 ≈ 0, the SFEE per unit mass is:

0 = (h2 − h1) + V2²/2
V2 = sqrt( 2·(h1 − h2) )

The enthalpy drop is 300 kJ/kg = 300,000 J/kg:

V2 = sqrt( 2 · 300,000 ) = sqrt(600,000) ≈ 775 m/s

The unit trap: enthalpy is in kJ/kg, so multiply by 1000 before taking the square root, or you’ll get 24.5 m/s instead of 775.

The kinetic-energy unit trap

h is kJ/kg; V²/2 is J/kg. One kJ = 1000 J. The nozzle calculator keeps the units straight — enter the inlet and exit states, toggle “neglect velocities” if the inlet velocity is small, and it returns the exit velocity in m/s (or ft/s in English units).

Throttling is isenthalpic

A valve or porous plug does no work, exchanges no heat, and has negligible velocity change, so h2 = h1. That single fact solves a huge class of refrigeration and steam problems: drop the pressure, keep the enthalpy, read the new state.