= Solution
The one-dimensional <mass flux> and <momentum flux> are respectively
$$
J_m=\rho u,\qquad J_p=p+\rho u^2.
$$
In a steady flow, the <conservation laws> make both the mass flux and the given <energy flux>
$$
J_E=u\left(p+W+\frac12\rho u^2\right)
$$
constant. Wherever $J_m\ne0$, their ratio is therefore constant:
$$
\frac{J_E}{J_m}
=\frac{p+W}{\rho}+\frac12u^2
=\frac{\gamma}{\gamma-1}\frac p\rho+\frac12u^2
=C,
$$
where $W=p/(\gamma-1)$ is the <internal energy> per unit volume of the <ideal gas>. This is the required steady-flow energy integral.
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