Crocco's equation

The momentum equation without body forces

$$\rho \frac{D𝐯}{Dt}=-\mathrm{\nabla }p$$

Expanding the substantial derivative

$$\rho \frac{\partial 𝐯}{\partial t}+\rho 𝐯\cdot \mathrm{\nabla }𝐯=-\mathrm{\nabla }p$$

The first and second law of thermodynamics gives

$$T\mathrm{\nabla }s=\mathrm{\nabla }h-\frac{\mathrm{\nabla }p}{\rho }$$

Insert $\mathrm{\nabla }p$ from the momentum equation

$$T\mathrm{\nabla }s=\mathrm{\nabla }h+\frac{\partial 𝐯}{\partial t}+𝐯\cdot \mathrm{\nabla }𝐯$$

Definition of total enthalpy ($h_o$)

$$h_o=h+\frac{1}{2}𝐯\cdot 𝐯\Rightarrow \mathrm{\nabla }h=\mathrm{\nabla }{h}_o-\mathrm{\nabla }(\frac{1}{2}𝐯\cdot 𝐯)$$

The last term can be rewritten as

$$\mathrm{\nabla }(\frac{1}{2}𝐯\cdot 𝐯)=𝐯\times (\mathrm{\nabla }\times 𝐯)+𝐯\cdot \mathrm{\nabla }𝐯$$

which gives

$$\mathrm{\nabla }h=\mathrm{\nabla }{h}_o-𝐯\times (\mathrm{\nabla }\times 𝐯)-𝐯\cdot \mathrm{\nabla }𝐯$$

Insert $\mathrm{\nabla }h$ in the entropy equation gives

$$T\mathrm{\nabla }s=\mathrm{\nabla }{h}_o-𝐯\times (\mathrm{\nabla }\times 𝐯)-\cancel{𝐯\cdot \mathrm{\nabla }𝐯}+\frac{\partial 𝐯}{\partial t}+\cancel{𝐯\cdot \mathrm{\nabla }𝐯}$$

$$T\mathrm{\nabla }s=\mathrm{\nabla }{h}_o-𝐯\times (\mathrm{\nabla }\times 𝐯)+\frac{\partial 𝐯}{\partial t}$$