The first law of thermodynamics applies to a material system: a fixed collection of matter. An open device such as a turbine, nozzle, heat exchanger, or filling tank is more naturally described by a control volume, where mass can cross the boundary.

For a material system, write the total energy as

$$E=U+K+E_g,$$

including internal, kinetic, and gravitational potential energy. With heat transfer into the system positive and work done by the system positive,

$$\boxed{\frac{dE_{\rm sys}}{dt}=\dot Q-\dot W}.$$

Reynolds transport theorem converts the system-energy rate into a fixed-control-volume balance.

Energy carried by flowing mass

Define the specific total energy excluding flow work as

$$e=u+\frac{V^2}{2}+gz,$$

where $u$ is specific internal energy.

If matter crosses a control surface against pressure, pressure forces do mechanical work that pushes fluid into or out of the control volume. For a small flowing volume $dV$, that pressure work is

$$p,dV.$$

Per unit mass, using specific volume $v=1/\rho$,

$$pv=\frac{p}{\rho}.$$

Combining this flow-work contribution with internal energy gives the specific enthalpy

$$\boxed{h=u+pv}.$$

Thus a flowing stream naturally transports

$$\boxed{h+\frac{V^2}{2}+gz}$$

per unit mass when pressure flow work is absorbed into enthalpy.

General fixed-control-volume balance

Separate shaft, electrical, and other non-flow work into $\dot W_s$. For a fixed control volume with discrete inlets and outlets,

$$\boxed{ \frac{dE_{CV}}{dt}

\dot Q-\dot W_s +\sum_{\rm in}\dot m\left(h+\frac{V^2}{2}+gz\right) -\sum_{\rm out}\dot m\left(h+\frac{V^2}{2}+gz\right) },$$

where

$$E_{CV}=\int_{CV}\rho\left(u+\frac{V^2}{2}+gz\right)dV.$$

The left side is energy accumulation inside the control volume. The terms on the right represent energy transferred by heat, non-flow work, and flowing matter.

Under this convention, $\dot Q>0$ means heat enters the control volume and $\dot W_s>0$ means shaft or other non-flow work leaves it.

Unsteady filling example

Consider a rigid insulated tank being filled from one inlet. There is no outlet, shaft work, or heat transfer:

$$\dot Q=0,\qquad \dot W_s=0.$$

Neglect gravitational-potential changes. If the entering stream has enthalpy $h_{in}$ and speed $V_{in}$,

$$\frac{dE_{CV}}{dt} =\dot m_{in}\left(h_{in}+\frac{V_{in}^2}{2}\right).$$

If inlet kinetic energy is negligible,

$$\boxed{\frac{dE_{CV}}{dt}=\dot m_{in}h_{in}}.$$

Suppose

$$\dot m_{in}=0.20,\mathrm{kg/s},\qquad h_{in}=300,\mathrm{kJ/kg}.$$

Then

$$\frac{dE_{CV}}{dt} =(0.20)(300) =\boxed{60,\mathrm{kW}}.$$

The tank gains energy even though the process is adiabatic because energy enters with the flowing mass.

Steady state as a special case

At steady state,

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

The general balance then reduces to the steady-flow energy equation. Further simplifications—one inlet and one outlet, negligible kinetic-energy change, adiabatic operation, and so on—must be justified for the device being modeled.

The control-volume energy balance is therefore the open-system form of the first law: energy can cross the boundary as heat, work, and energy carried by mass.