 Solution

For a barotropic fluid, define $h(\rho )=\int dp/\rho$. Dotting the steady Euler equation with $\mathbf{u}$ givesThus the Bernoulli function $H=u^2/2+h+\psi$ is constant along each streamline. Different streamlines can have different constants; steady flow alone does not imply one spatially uniform $H$.

In a quasi-one-dimensional flow through area $A(z)$ with negligible gravity, the axial momentum equation and $dp=c_s^{2}d\rho$ giveMeanwhile the mass flow rate $\dot{M}=\rho uA$ is constant, so $d\ln \rho =-d\ln u-d\ln A$. Substitution proves the nozzle area-velocity relationA subsonic flow accelerates where $A$ decreases, whereas a supersonic flow accelerates where $A$ increases. A smooth accelerating transonic branch in a de Laval nozzle therefore crosses $u=c_s$ at a minimum-area throat. Suitable reservoir and downstream pressures are also required: a converging-diverging shape by itself does not exclude a wholly subsonic solution or a downstream shock.

For the stipulated constant-$K$ polytropic flow, $c_s^2=\gamma K{\rho }^{\gamma -1}$. At the throat $u_m=c_{s,m}$, soThe specific enthalpy is $h=\gamma K{\rho }^{\gamma -1}/(\gamma -1)=c_s^2/(\gamma -1)$. Evaluating $H$ at the throat and substituting $\rho =\dot{M}/(Au)$ elsewhere givesOn the outward supersonic branch, $A\sim C{r}^2$ lets the gas expand and its specific enthalpy tend to zero for $\gamma >1$. ConsequentlyFor $\gamma =5/3$, $u_{\infty }=2c_{s,m}$. Also $T\propto {\rho }^{\gamma -1}\propto r^{-2(\gamma -1)}$. These are the adiabatic nozzle terminal speed and expansion laws for a lossless escaping branch; a subsonic branch could instead approach finite pressure with $u\propto A^{-1}$. A pressure-shaped astrophysical jet is being approximated by a nozzle here: the assumed geometry does not by itself explain the confinement or account for magnetic forces or relativistic motion in an actual active galactic nucleus.