= Solution
The steady <Euler equations for an inviscid fluid> are
$$
\rho(u\cdot\nabla)u=-\nabla p-\nabla\chi.
$$
Taking the <scalar> product with $u$ and using
$$
u\cdot(u\cdot\nabla u)
=u\cdot\nabla\left(\frac12|u|^2\right)
$$
gives
$$
u\cdot\nabla
\left(\frac12\rho|u|^2+p+\chi\right)=0.
$$
Thus
$$
\boxed{u\cdot\nabla H=0}.
$$
The <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 gives
$$
aU=A\dot y.
$$
The 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, gives
$$
\frac12\left(U^2-\dot y^2\right)
=g(H+h_0-y).
$$
Therefore
$$
\dot y=
\sqrt{\frac{2g(H+h_0-y)}{A^2/a^2-1}}.
$$
The surface reaches the upper tube end when $y=h_0$. Hence the <draining time of a uniform tank through a siphon> is
$$
\begin{aligned}
t
&=\sqrt{\frac{A^2/a^2-1}{2g}}
\int_0^{h_0}\frac{dy}{\sqrt{H+h_0-y}}\\
&=\boxed{
\sqrt{2}\left(\frac{A^2}{a^2}-1\right)^{1/2}
\frac{\sqrt{H+h_0}-\sqrt H}{\sqrt g}}.
\end{aligned}
$$
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