2.3.1 Basic thermodynamic equations

Differential relations:

$${\rm d} e = T {\rm d} s + \frac{p}{\rho^2} {\rm d} \rho$$ (16)

where $e$ is the internal energy .

$${\rm d} h = T {\rm d} s + \frac{1}{\rho} {\rm d} p$$ (17)

where the specific enthalpy , $h$, is defined as

$$h = e + p/\rho$$ (18)

This implies:

$$\left(\frac{\partial e}{\partial s}\right)_{\rho} = T$$ (19)

$$\left(\frac{\partial e}{\partial \rho}\right)_{s} = \frac{p}{\rho^2}$$ (20)

$$\left(\frac{\partial h}{\partial s}\right)_{p} = T$$ (21)

$$\left(\frac{\partial h}{\partial p}\right)_{s} = \frac{1}{\rho}$$ (22)