First Law of Thermodynamics for Closed Systems

The energy in a closed system, where there is no mass flow across the system boundary, is balanced such that:

$$\begin{array}{rl}\Delta E_{\text{system}}& =E_{\text{in}}-E_{\text{out}}\end{array}$$

In rate form:

$${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=d{E}_{\text{system}}/dt$$

The total energy of the system, $E_{\text{system}}$ is the sum of 3 components:

$$E_{\text{system}}=U+KE+PE$$

Often, we only consider the internal energy, $\Delta E_{\text{system}}=\Delta U$, since much of the time the system is stationary, such that $\Delta KE=\Delta PE=0$. In this case, we write:

$$\Delta U=Q-W$$

The sign convention is to assume that heat, $Q$, is being transferred into the system and work $W$ is done by the system.

Note that in the full form with kinetic and potential energy, we have:

$$\begin{array}{rl}\Delta \text{KE}+\Delta \text{PE}+\Delta U_{\text{system}}& =Q-W\\ \frac{d(\text{KE})}{dt}+\frac{d(\text{PE})}{dt}+\frac{dU}{dt}& =\dot{Q}-\dot{W}\end{array}$$

where the left side is the rate at which system energy changed, $\dot{Q}$ is the rate of heat addition, and $\dot{W}$ is the rate of work done (power).

Steady-state

At a steady state such that $d{E}_{\text{CV}}/dt=0$, we have

$$0=\dot{Q}-\dot{W}+{\dot{m}}_{\text{in}}{\left(h+\frac{1}{2}{V}^2+gz\right)}_{\text{in}}-{\dot{m}}_{\text{out}}{\left(h+\frac{1}{2}{V}^2+gz\right)}_{\text{out}}$$

One inlet, one outlet and steady

Extrinsic form

$$0=\dot{Q}-\dot{V}+\dot{m}\left[{\left(h+\frac{1}{2}{V}^2+gz\right)}_{\text{in}}-{\left(h+\frac{1}{2}{V}^2+gz\right)}_{\text{out}}\right]$$

Intrinsic form