For a fixed control volume operating at steady state, the energy stored inside the control volume does not change with time:

$$\frac{dE_{CV}}{dt}=0.$$

The general control-volume first law therefore reduces to a balance among heat transfer, non-flow work, and energy carried by inlet and outlet streams.

For one inlet and one outlet with equal steady mass flow rate $\dot m$,

$$\boxed{ \dot Q-\dot W_s =\dot m\left[ (h_2-h_1) +\frac{V_2^2-V_1^2}{2} +g(z_2-z_1) \right] },$$

where $h$ is specific enthalpy, $V^2/2$ is specific kinetic energy, $gz$ is specific gravitational potential energy, $\dot Q$ is heat-transfer rate into the control volume, and $\dot W_s$ is shaft or other non-flow power delivered by the control volume under this sign convention.

What steady state means

Steady state does not mean that no energy passes through the device. It means that macroscopic properties at fixed locations and the total energy stored inside the control volume do not change with time.

A turbine can continuously receive energy in a hot flowing stream and continuously deliver shaft power while remaining at steady state.

Simplify only after writing the full balance

Different devices become simple because different terms are negligible—not because each device obeys a different energy law.

For example:

• an adiabatic device has $\dot Q\approx0$;
• a device with negligible elevation change has $z_2\approx z_1$;
• a large reservoir or slow pipe flow can have negligible kinetic-energy change;
• a nozzle can have negligible shaft work while converting enthalpy into kinetic energy.

These are modeling assumptions and must be justified for the particular device.

Worked example: adiabatic turbine

A turbine operates steadily with

$$\dot m=3.0,\mathrm{kg/s}$$

and its specific enthalpy decreases by

$$h_1-h_2=120,\mathrm{kJ/kg}.$$

Assume the turbine is adiabatic and inlet/outlet kinetic and potential energy changes are negligible. Then

$$0-\dot W_s=\dot m(h_2-h_1),$$

so

$$\dot W_s =\dot m(h_1-h_2) =(3.0)(120) =\boxed{360,\mathrm{kW}}.$$

Worked comparison: ideal nozzle

For a steady adiabatic nozzle with no shaft work and negligible elevation change,

$$0=(h_2-h_1)+\frac{V_2^2-V_1^2}{2}.$$

Thus

$$\boxed{h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}}.$$

A decrease in enthalpy can therefore produce an increase in flow speed.

The SFEE is not a separate conservation principle. It is the steady one-inlet/one-outlet specialization of the general control-volume energy balance.