A strictly conservative form of the Navier-Stokes energy equation reads (Schlichting, Gersten: Boundary Layer Theory 2017, eq. 3.57) $$\frac{\partial \rho e_t}{\partial t} + \mathrm{div}\left(\rho e_t \mathbf{u}\right) = \mathrm{div}\left(\mathbf{\sigma}\cdot \mathbf{u}\right) - \mathrm{div}\left(\mathbf{q}\right) + Q$$ where $\rho$ is density, $e_t = e + \frac{1}{2} \mathbf{u}^T \cdot \mathbf{u} + \psi$ is the specific total energy, $e$ is the specific internal energy, $\mathbf{u}$ is the flow velocity, $\psi$ is the potential energy, $\mathbf{\sigma} = \mathbf{\tau} - p \mathrm{\mathbf{I}}$ is the Cauchy stress tensor, $\mathbf{\tau}$ is the viscous stress tensor, $p$ is the pressure, $\mathrm{\mathbf{I}}$ is the second-order identity tensor, $\mathbf{q}$ is the heat conduction vector and $Q$ is a point-like heat source.
In terms of total enthalpy $h_t = e_t + p/\rho$, this equation reads $$\frac{\partial \rho h_t}{\partial t} + \mathrm{div}\left(\rho h_t \mathbf{u}\right) - \frac{\partial p}{\partial t} - \mathrm{div}\left(p \mathbf{u}\right) = \mathrm{div}\left(\mathbf{\sigma}\cdot \mathbf{u}\right) - \mathrm{div}\left(\mathbf{q}\right) + Q$$ or (with $\mathbf{\sigma} = \mathbf{\tau} - p \mathrm{\mathbf{I}}$) $$\frac{\partial \rho h_t}{\partial t} + \mathrm{div}\left(\rho h_t \mathbf{u}\right) - \frac{\partial p}{\partial t} = \mathrm{div}\left(\mathbf{\tau}\cdot \mathbf{u}\right) - \mathrm{div}\left(\mathbf{q}\right) + Q$$
I always thought total enthalpy was the sum of all energy forms that are connected to mass. Thus, total enthalpy can be transported/convected. So why is the pressure-volume work $pv = p/\rho$ not included? In my opinion, the energy equation should read $$\frac{\partial \rho h_t}{\partial t} + \mathrm{div}\left(\rho h_t \mathbf{u}\right) = \mathrm{div}\left(\mathbf{\sigma}\cdot \mathbf{u}\right) - \mathrm{div}\left(\mathbf{q}\right) + Q$$ Where is my error in reasoning? Even Bernoulli's equation for incompressible flow includes the term $p/\rho$.
