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Uniformly stratified, nonrotating, two-dimensional ideal Boussinesq equations in coordinates $(x,z)$ are equivalent to unstratified rotating equations independent of $Y$, with $(X,Y,Z)=(z,y,-x)$, rotation $2\mathbf{\Omega }=N{\mathbf{e}}_Z$, velocity $(U,V,W)=(w,\sigma /N,-u)$, and the same reduced pressure. The transverse rotating velocity represents buoyancy perturbation, while the rotation-axis velocity represents the stratified horizontal velocity. Matching prescribed forces may have components in the invariant plane, but no transverse component. Physical free surface conditions are not preserved: the physical vertical direction and its normal velocity are different under the mapping.

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.

A uniform steady film of thickness $d$ on a plane inclined by $\theta$ above horizontal has downslope coordinate $s$ and outward normal coordinate $x$. Neglecting air shear, the pressure is $p_{\mathrm{atm}}+\rho g\cos \theta (d-x)$ and its downslope gradient vanishes. The Newtonian fluid momentum equation is $\eta v^{\prime \prime }+\rho g\sin \theta =0$. The no-slip boundary condition $v(0)=0$ and shear-free free surface $v^{\prime }(d)=0$ give the displayed parabolic profile. Integrating over thickness gives volume flux per unit width $q$. The wall shear $\rho gd\sin \theta$ balances the film's downslope weight, and the vertical-wall case is $\theta =\pi /2$.

For a fixed half-angle $0<\alpha <\pi /2$, a slow slender gravity current in a V-shaped channel has an almost horizontal free surface across each section. Its area is $h^2\tan \alpha$, and its axial volume flux is $-gQ(\alpha )h^4{h}_x/\nu$. Here $Q$ is the integral of the dimensionless cross-sectional Poisson equation solution: $\Delta \varphi =-1$, with zero velocity on the two rigid slopes and zero normal velocity derivative on the horizontal shear-free surface. Conservation of mass gives $C=gQ/(\nu \tan \alpha )$. This treats the full two-dimensional section and does not require a shallow V angle.