View source for Crocco's equation

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The momentum equation without body forces

{{NumEqn|<math>
\rho\frac{D\mathbf{v}}{Dt}=-\nabla p
</math>}}

Expanding the substantial derivative

{{NumEqn|<math>
\rho\frac{\partial \mathbf{v}}{\partial t}+\rho\mathbf{v}\cdot\nabla\mathbf{v}=-\nabla p
</math>}}

The first and second law of thermodynamics gives

{{NumEqn|<math>
T\nabla s =\nabla h-\frac{\nabla p}{\rho}
</math>}}

Insert <math>\nabla p</math> from the momentum equation

{{NumEqn|<math>
T\nabla s =\nabla h+\frac{\partial \mathbf{v}}{\partial t}+\mathbf{v}\cdot\nabla\mathbf{v}
</math>}}

Definition of total enthalpy (<math>h_o</math>)

{{NumEqn|<math>
h_o=h+\frac{1}{2}\mathbf{v}\cdot\mathbf{v}\Rightarrow \nabla h=\nabla h_o-\nabla\left(\frac{1}{2}\mathbf{v}\cdot\mathbf{v}\right)
</math>}}

The last term can be rewritten as

{{NumEqn|<math>
\nabla\left(\frac{1}{2}\mathbf{v}\cdot\mathbf{v}\right)=\mathbf{v}\times(\nabla\times\mathbf{v})+\mathbf{v}\cdot\nabla\mathbf{v}
</math>}}

which gives

{{NumEqn|<math>
\nabla h=\nabla h_o-\mathbf{v}\times(\nabla\times\mathbf{v})-\mathbf{v}\cdot\nabla\mathbf{v}
</math>}}

Insert <math>\nabla h</math> in the entropy equation gives

{{NumEqn|<math>
T\nabla s =\nabla h_o-\mathbf{v}\times(\nabla\times\mathbf{v})-\cancel{\mathbf{v}\cdot\nabla\mathbf{v}}+\frac{\partial \mathbf{v}}{\partial t}+\cancel{\mathbf{v}\cdot\nabla\mathbf{v}}
</math>}}

{{NumEqn|<math>
T\nabla s =\nabla h_o-\mathbf{v}\times(\nabla\times\mathbf{v})+\frac{\partial \mathbf{v}}{\partial t}
</math>}}