Crocco's equation

The momentum equation without body forces

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

(Eq. 2.86)

Expanding the substantial derivative

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

(Eq. 2.87)

The first and second law of thermodynamics gives

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

(Eq. 2.88)

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

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

(Eq. 2.89)

Definition of total enthalpy ($h_o$)

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

(Eq. 2.90)

The last term can be rewritten as

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

(Eq. 2.91)

which gives

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

(Eq. 2.92)

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

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

(Eq. 2.93)

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

(Eq. 2.94)