= Draining tank with constant cross-sectional area
{title2=$h(t)=H(1-t/T)^2$}
Take a reservoir with constant horizontal cross-sectional <area> $A$, one small <orifice> of <area> $a$, no inflow, and constant stipulated <discharge coefficient> $C_d>0$. Let the initial head be $H>0$. For constant <mass density> $\rho$, the stored <mass> above the outlet level is $\rho Ah$ and outgoing <mass flow rate> is $\rho Q$. <Mass conservation> cancels $\rho$ and gives the <initial value problem>
$$
A\frac{dh}{dt}=-Q,
\qquad \frac{dh}{dt}=-k\sqrt h,
\qquad h(0)=H,
\qquad k=\frac{C_da}{A}\sqrt{2g}.
$$
The downward <speed> of the <free surface> is $Q/A$, smaller than the ideal jet <speed> by the factor $C_da/A$. Thus a small opening relative to the reservoir supports the negligible-surface-<velocity> assumption.
For $h>0$, <separation of variables> or the <chain rule> gives $d\sqrt h/dt=-k/2$. Integrating yields
$$
T=\frac{2\sqrt H}{k}=\frac{A}{C_da}\sqrt{\frac{2H}{g}},
\qquad
h(t)=\begin{cases}H(1-t/T)^2,&0\le t\le T,\\0,&t\ge T,\end{cases}
$$
$$
Q(t)=\begin{cases}Q_0(1-t/T),&0\le t\le T,\\0,&t\ge T,\end{cases}
\qquad Q_0=C_da\sqrt{2gH}.
$$
The model reaches zero head in finite time even though <volumetric flow rate> tends to zero. Extending the squared expression beyond $T$ would make head grow again and violate the original <ordinary differential equation>. The zero-head continuation enforces no further outflow. For an elevated side opening, zero head means the surface has reached that opening, leaving <fluid> below it; $T$ is not the time to remove every drop from such a tank.
For an original synthetic example, stipulate
$$
A=0.100\ \mathrm{m^2},\quad a=1.00\times10^{-4}\ \mathrm{m^2},\quad
C_d=0.620,\quad H=0.250\ \mathrm m,\quad g=9.81\ \mathrm{m\,s^{-2}}.
$$
The initial <volume> above outlet level is $AH=0.0250\ \mathrm{m^3}=25.0\ \mathrm L$. The initial ideal jet <speed> is $2.215\ \mathrm{m\,s^{-1}}$, the initial <volumetric flow rate> is $Q_0=0.1373\ \mathrm{L\,s^{-1}}$, and the predicted zero-head time is $T=364.1\ \mathrm s$. These are consequences of stipulated parameters, not experimental measurements. At $t=T/2=182.1\ \mathrm s$, head is $H/4=0.0625\ \mathrm m$, stored <volume> is $6.25\ \mathrm L$, and <volumetric flow rate> is $Q_0/2=0.06866\ \mathrm{L\,s^{-1}}$. Half the initial <volume> has already left at $t=T(1-1/\sqrt2)=106.7\ \mathrm s$.
\Image[/torricelli-draining-tank-height-and-flow.png]
{title=Constant-area drainage: quadratic head and linearly declining discharge}
{description=Original synthetic trajectories for the stipulated <draining tank with constant cross-sectional area>, with $A=0.100\ \mathrm{m^2}$, $a=10^{-4}\ \mathrm{m^2}$, $C_d=0.620$, $H=0.250\ \mathrm m$ and $g=9.81\ \mathrm{m\,s^{-2}}$. The model remains at zero head and zero <volumetric flow rate> after $T$. The shaded <area> under the flow curve represents $25.0\ \mathrm L$.}
An independent <mass conservation> check integrates the outgoing <volumetric flow rate>:
$$
\int_0^tQ(s)\,ds
=Q_0\left(t-\frac{t^2}{2T}\right)
=A[H-h(t)]\quad(0\le t\le T),
\qquad \int_0^TQ(s)\,ds=\frac{Q_0T}{2}=AH.
$$
Multiply both sides by <mass density> to obtain discharged <mass>. A <dimensional analysis> check gives $[k]=\mathrm{m^{1/2}\,s^{-1}}$, $[Q]=\mathrm{m^3\,s^{-1}}$ and $[T]=\mathrm s$. The triangular <area> under $Q(t)$ explains why the initial-rate estimate $AH/Q_0$ is only half the actual model time. Doubling $A$ doubles $T$; doubling $a$ or $C_d$ halves $T$, within the same assumptions.
This finite-time result is an idealized prediction. The constant $C_d$ and <quasi-steady approximation> need not survive very small head or the initial startup. Appreciable <viscosity> in a narrow outlet tube requires a different <pressure> balance, as in <Poiseuille flow>; appreciable <surface tension> can prevent zero head altogether.
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