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The steady Bernoulli equation isAt the regular sonic point, part a givesDefineThe isentropic flow relation $c_s^2\propto {\rho }^{\gamma -1}$ and mass conservation between the stellar surface and the sonic point giveEquating the surface and sonic values of the Bernoulli function therefore yields the required relationAn outflow reaching infinity with positive terminal kinetic energy requires $E>0$, henceAt $\gamma =5/3$ the terminal state is marginal with $E=0$; larger $\gamma$ cannot support the stipulated wind to infinity.

Let $u=-v_r>0$ denote the inward speed. Steady spherical mass conservation and the radial momentum equation for an isothermal gas giveEliminating $d\log \rho /dr$ produces the Isothermal Bondi equationA smooth flow can cross Mach number one only where both sides vanish. Its Bondi sonic point is thereforeThe integrated Bernoulli equation which approaches rest and density ${\rho }_{\infty }$ at infinity isAt the sonic point this gives ${\rho }_s=e^{3/2}{\rho }_{\infty }$, and hence the unique regular transonic branch has

The steady Euler equations for an inviscid fluid areTaking the scalar product with $u$ and usinggivesThusThe Bernoulli function $H$ is constant along each streamline: fluid particles in steady flow exchange pressure, kinetic, and potential energy without changing their total mechanical energy density.

Let $y(t)$ be the downward displacement of the free surface from its initial level, and let $U$ be the speed in the tube. Conservation of volume givesThe free surface and the outlet are both at atmospheric pressure, and their vertical separation is $H+h_0-y$. Applying the Bernoulli equation between them, while retaining the small free-surface speed, givesThereforeThe surface reaches the upper tube end when $y=h_0$. Hence the draining time of a uniform tank through a siphon is

Define the indoor-to-outdoor reduced gravityBecause the whole opening lies above the interface, its indoor side contains upper-layer fluid of density ${\rho }_H$. Under hydrostatic pressure, the pressure difference varies linearly about a neutral pressure level. Equal opening geometry and equal discharge coefficients for inflow and outflow put that level at the vent midpoint. At vertical distance $y$ from it, the ideal Bernoulli equation gives speed $\sqrt{2g_V^{\prime }|y|}$.

Writing $Q_V$ for either one-way volumetric flow rate, integration over one half of the opening givesThe total unsigned exchange is $2Q_V$. The ideal sharp-edged inviscid model has $C_d=1$; an empirical $C_d<1$ represents contraction and losses.

Uniform irrotational flow of speed $U$ past a circular cylinder of radius $a$, with zero circulation, has velocity potentialOn the cylinder, the tangential speed is $-2U\sin \theta$, so the Bernoulli equation gives

For steady spherical isentropic flow with $p=K{\rho }^{3/2}$ around a point mass, put $j=r^2\rho v>0$ and $\mathcal{E}=v^2/2+2c^2-GM/r>0$. Eliminating the mass density gives $c^2=[j(3K/2{)}^2{]}^{2/5}{r}^{-4/5}{\mathcal{M}}^{-2/5}$. With $r_s=GM/(4\mathcal{E})$, $x=(r/r_s{)}^{1/5}$, $y={\mathcal{M}}^{2/5}$ and $\lambda =[j(3K/2{)}^2{]}^{2/5}/(2\mathcal{E}{r}_s^{4/5})$, the Bernoulli equation reduces to $\lambda F(y)=F(x)$, where $F(t)=t^4+4/t$. Since $F$ has its unique minimum at $t=1$, a regular sonic point requires $\lambda =1$ and $r=r_s$. One transonic branch is $y=x$, which has $v=\sqrt{2\mathcal{E}}$, $\rho \propto r^{-2}$ and $p\propto r^{-3}$. The decreasing branch has inner free fall and an outer nearly static reservoir, giving Bondi accretion for inward fluid flow. A nonradiative stationary shock wave keeps $j,\mathcal{E},r_s$ fixed but raises $K$, so $\lambda$ increases by the factor $(K_2/K_1{)}^{4/5}$. The inward transonic solution is the spherical $n=2$, $\gamma =3/2$ specialization of transonic accretion in a power-law tube.

Consider steady isentropic flow toward a Newtonian gravitational potential $-GM/r$ through a tube of cross-sectional area $A(r)=C{r}^n$. Write $v>0$ for the inward speed and use a polytropic equation of state with specific-heat ratio $\gamma >1$. Mass conservation and the Euler equations for an inviscid fluid givewhere $c_s$ is the adiabatic sound speed. A regular sonic point therefore has $v_s=c_s$ and $r_s=GM/(n{c}_s^2)$. The Bernoulli equation, matched to a nearly stationary reservoir with sound speed $c_0$, givesFor $n>1/2$, a finite positive sonic point requires $1<\gamma <(2n+1)/(2n-1)$. Differentiating the flow equation at that point, with $x=r_s{v}_s^{\prime }/c_s$, givesIts discriminant is $4n[(2n+1)-(2n-1)\gamma ]$, so the same bound permits real regular slopes. The branch on which the Mach number rises inward selects the negative sign inFor a spherical tube $n=2$, this recovers the $5/3$ threshold of Bondi accretion; for a dipolar flux-tube area, $n=3$ gives $7/5$. The tube approximation must remain valid between the reservoir matching region and the accretor.