Editing Euler equations (fluid dynamics) (section)

===Compressible case ===
In the most general steady (compressible) case the mass equation in conservation form is:

<math display="block"> \nabla \cdot \mathbf j = \rho \nabla \cdot \mathbf u + \mathbf u \cdot \nabla \rho = 0.</math>Therefore, the previous expression is rather

<math display="block">\mathbf{u} \cdot \nabla \left({\frac{1}{2}}u^2 + \phi + \frac{p}{\rho}\right) = \frac{p}{\rho}\nabla \cdot \mathbf{u}.</math>

The right-hand side appears on the energy equation in convective form, which on the steady state reads:

<math display="block">\mathbf u \cdot \nabla  e =  - \frac{p}{\rho}  \nabla \cdot \mathbf u. </math>

The energy equation therefore becomes:

<math display="block">\mathbf u \cdot \nabla \left( e + \frac p \rho + \frac 1 2 u^2 + \phi \right) = 0, </math>

so that the internal specific energy now features in the head.

Since the external field potential is usually small compared to the other terms, it is convenient to group the latter ones in the [[total enthalpy]]:

<math display="block"> h^t \equiv e + \frac p \rho + \frac 1 2 u^2,</math>

and the [[Bernoulli invariant]] for an inviscid gas flow is:

<math display="block"> b_g \equiv h^t + \phi = b_l + e , </math>

which can be written as:

<math display="block">\mathbf u \cdot \nabla b_g = 0.</math>

That is, '''the energy balance for a steady inviscid flow in an external conservative field states that the sum of the total enthalpy and the external potential is constant along a streamline'''.

In the usual case of small potential field, simply:

<math display="block">\mathbf u \cdot \nabla h^t \sim 0.</math>