Exergy

Remarks

Kinetic and potential energy of the system as a whole, when present, are added to $X$ in full. For a steady flow, the specific flow exergy is $(h - h_0) - T_0 (s - s_0)$ plus kinetic and potential terms, where $h$ is the specific enthalpy and $s$ the specific entropy.

Example: a body of constant heat capacity $C$ at temperature $T$, with negligible volume change, has $U - U_0 = C(T - T_0)$ and $S - S_0 = C \ln(T / T_0)$. So $X = C\big(T - T_0 - T_0 \ln(T / T_0)\big) = C T_0 (y - 1 - \ln y)$ with $y = T / T_0$. Since $\ln y < y - 1$ for $y \ne 1$, the exergy is positive whenever $T \ne T_0$, both for a hot and for a cold body.

Non-example: the internal energy $U$ alone is not the exergy. A system already in the dead state has $X = 0$ although $U_0$ may be large, because no work can be extracted from it with that environment.

By the second law, $X \ge 0$, and $X = 0$ in the dead state.