= Solution
For a <barotropic fluid>, define $h(\rho)=\int dp/\rho$. Dotting the steady <Euler equation> with $\mathbf u$ gives
$$
\mathbf u\cdot\nabla\left(\frac{u^2}{2}+h+\psi\right)=0.
$$
Thus 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 $dp=c_s^2d\rho$ give
$$
u\frac{du}{dz}=-\frac{c_s^2}{\rho}\frac{d\rho}{dz},\qquad
u^2\frac{d\ln u}{dz}=-c_s^2\frac{d\ln\rho}{dz}.
$$
Meanwhile 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 relation>
$$
\boxed{(u^2-c_s^2)\,d\ln u=c_s^2\,d\ln A.}
$$
A <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>.
\Image[/past-exam-of-the-astrophysical-fluid-dynamics-course-of-the-university-of-cambridge/2013/paper-1-de-laval-nozzle.png]
{title=Choked adiabatic nozzle: minimum area, sonic crossing and continued supersonic acceleration}
For the stipulated constant-$K$ <polytropic flow>, $c_s^2=\gamma K\rho^{\gamma-1}$. At the throat $u_m=c_{s,m}$, so
$$
\dot M=A_m\rho_m\sqrt{\gamma K\rho_m^{\gamma-1}},\qquad
\boxed{\rho_m=\left(\frac{\dot M}{A_m\sqrt{\gamma K}}\right)^{2/(\gamma+1)}.}
$$
The <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 gives
$$
\boxed{\frac{u^2}{2}+\frac{\gamma K}{\gamma-1}\left(\frac{\dot M}{Au}\right)^{\gamma-1}
=\frac{\gamma+1}{2(\gamma-1)}c_{s,m}^2.}
$$
On the outward supersonic branch, $A\sim Cr^2$ lets the gas expand and its <specific enthalpy> tend to zero for $\gamma>1$. Consequently
$$
\boxed{u_\infty=\sqrt{\frac{\gamma+1}{\gamma-1}}\,c_{s,m},\qquad
\rho\sim\frac{\dot M}{Cu_\infty}\,r^{-2}.}
$$
For $\gamma=5/3$, $u_\infty=2c_{s,m}$. 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>.
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