= Torricelli's law
{c}
{title2=$v=\sqrt{2gh}$}
{wiki}
An ideal fluid issuing from a small hole a vertical depth $h$ below a free surface has speed
$$
v=\sqrt{2gh}.
$$
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>.<Bernoulli’s Equation by NASA Glenn Research Center>[Macro Scale Derivation and Common Assumptions: steady, inviscid, incompressible streamline balance and the gravity-height term]{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> gives
$$
\frac{p_{\mathrm{ambient}}}{\rho}+gz_s
=\frac{p_{\mathrm{ambient}}}{\rho}+\frac{v^2}{2}+gz_o,
\qquad h=z_s-z_o.
$$
Equal <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.
\Image[/torricelli-reservoir-head-and-pressure.png]
{title=Torricelli geometry: equal boundary pressures and a vertical head}
{description=Original schematic of ideal outflow, not measured apparatus or a scale drawing. The <free surface> and emerging jet share ambient <pressure>. Head $h$ is measured to the <orifice> centre. Bulk <hydrostatic pressure> at that elevation is a different quantity from jet <pressure>.}
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>:
$$
Q=C_d a\sqrt{2gh},\qquad C_d=\frac{Q}{a\sqrt{2gh}},
$$
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 $Q=a_jv_j$ gives $C_d=C_cC_v$. 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.
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