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For a cone of half angle $\pi /4$ draining through a small circular aperture of radius $ϵ$, the horizontal area at height $h$ is $\pi h^2$. Torricelli's law gives speed $\sqrt{2gh}$, so conservation of mass gives $h^2\dot{h}=-{ϵ}^2\sqrt{2gh}$. Integration yields the displayed leading drainage time when $h_0\gg ϵ$. The final layer $h=O(ϵ)$ is outside the quasi-steady approximation and contributes only order $\sqrt{ϵ/g}$, compared with the leading time of order $h_0^{5/2}/({ϵ}^2\sqrt{g})$. This is an ideal inviscid estimate without a discharge coefficient.

A physicist and mathematician associated here with Torricelli's law.^[ref]

A physical oceanographer and author of the tank-draining discussion cited in Torricelli's law.^[ref][ref]

For constant fluid density $\rho$, steady Euler equations for an inviscid fluid under a conservative force per unit mass $-\nabla \Phi$ read $(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla p/\rho -\nabla \Phi$. Here $\mathbf{u}$ is velocity, $p$ pressure and $\Phi$ potential per unit mass. Dotting with $\mathbf{u}$ givesThis is the Bernoulli equation. The constants can differ between streamlines. If density varies, the same derivation for a barotropic fluid replaces $p/\rho$ by ${\int }^{p}d{p}^{\prime }/\rho (p^{\prime })$; constant density is the appropriate water approximation here.

For the draining cone, the radius at height $h$ is $h$ and the horizontal area is $\pi h^2$. Both the free surface and the jet are at atmospheric pressure; their gravitational potential difference is approximately $gh$. The speed of the descending surface is smaller than the jet speed by the area ratio ${ϵ}^2/h^2$, so it can be neglected when $h\gg ϵ$. The Bernoulli equation then gives Torricelli's law, $U\simeq \sqrt{2gh}$. Conservation of mass givesHence the leading drainage time of a conical tank isRemoving the tip places the aperture at height of order $ϵ$, and the quasi-steady calculation fails in the last layer of that depth. Integrating down only to $h=O(ϵ)$ changes the displayed leading time by relative order $(ϵ/h_0{)}^{5/2}$; replacing the actual head $h-ϵ$ by $h$ introduces a larger, but still vanishing, relative correction of order $ϵ/h_0$. The formula is the ideal inviscid estimate under the question's stated approximation, not an exact law for the final transient or a real contracted jet.

If $z(t)$ is the water depth and $A(z)=\pi r(z{)}^2$ is the horizontal cross-sectional area, Torricelli's law gives the volume flux through the hole asConservation of volume givesFor a prescribed constant fall rate $\dot{z}=-\alpha$, $\alpha >0$, this requiresConsequentlyso the container radius must be proportional to the fourth root of height.

In steady state the incoming volumetric flow rate equals the discharge. Bernoulli's principle between a nearly stationary free surface and the atmospheric outlet gives the ideal Torricelli's law speed $\sqrt{2gh}$. The specified discharge coefficient multiplies the opening-flow rate, soThus, with $A=10^{-4}{\mathrm{m}}^2$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$,Neglect surface-speed and capillary corrections. The coefficient includes contraction and loss effects; it should not be applied a second time to the calculated height.

An ideal fluid issuing from a small hole a vertical depth $h$ below a free surface has speed

Here $h$ is the vertical distance from the free surface to the centre of the orifice, and $g$ is the positive magnitude of gravitational acceleration. The ideal prediction assumes incompressible flow, the same ambient pressure at the free surface and emerging jet, negligible velocity of the reservoir's free surface, and negligible viscous dissipation and surface tension effects. For a falling free surface, a quasi-steady approximation additionally assumes that the local outflow adjusts much faster than $h$ changes. It applies the Bernoulli equation at each instant, rather than claiming that the whole drainage is steady flow.^[ref]

A short proof uses a streamline connecting the reservoir's free surface at elevation $z_s$ to the emerging jet at elevation $z_o$. With constant mass density $\rho$, negligible surface speed and no mechanical energy losses, the Bernoulli equation givesEqual pressures cancel, leaving $v^2/2=gh$: a unit mass gains the same kinetic energy as a body in free fall through height $h$. This makes Torricelli's law a useful first estimate before the detailed velocity field is known. At one quarter of the original head, ideal jet speed is one half as large; it is not proportional to head.

Far from the accelerating outlet region, the reservoir has approximately hydrostatic pressure $p_{\mathrm{ambient}}+\rho gh$ at the outlet's elevation. The emerging jet instead has approximately ambient pressure when surface tension is negligible. Do not assign that bulk hydrostatic pressure to the emerging jet as well as adding the height difference: that would count the available mechanical energy twice. A sealed reservoir, a submerged outlet or significant surface tension changes the pressure conditions and requires a different balance.

Actual volumetric flow rate is commonly represented by a discharge coefficient:where $a$ is the geometric area of the orifice. This definition compares volumetric flow rates, not just speeds. If a nearly uniform jet at its vena contracta has area $a_j$ and speed $v_j$, define the two separate ratios $C_c=a_j/a$ and $C_v=v_j/\sqrt{2gh}$. Then $Q=a_j{v}_j$ gives $C_d=C_c{C}_v$. A contracted jet can consequently have a speed close to $\sqrt{2gh}$ while $C_d$ is appreciably below one. No particular $C_d$ is universal: calibrate it for the opening and operating range.