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James F. Price's Draining a Tank, an appendix to Dimensional Analysis of Models and Data Sets, describes an actual tabletop experiment in sections 2.1–2.3. Tap water filled a semitransparent polypropylene beaker with a marked volume scale and drilled side openings. Some trials added horizontal outlet tubes. Temperature was measured, and dye checked that the water had settled before uncovering the outlet. The readout was elapsed time at successive volume marks. Opening diameter and tube length varied between trials. Reported repeatability was checked; tube bores were measured with feeler gauges rather than accepted from nominal sizes.^[ref]

For tank area $A$ and volume $W_o$ at outlet level, the observations give $h_i=(W_i-W_o)/A$ and interval volumetric flow rate $Q_i\simeq (W_i-W_{i+1})/(t_{i+1}-t_i)$. Price compared $Q/(a\sqrt{2gh})$ across heads to infer a discharge coefficient where that ratio was nearly constant. At small head, drainage could stop with a protruding water cap; surface tension mattered. Outlet tubes introduced viscosity effects. This experiment tests a useful operating range, rather than proving ideal drainage down to zero head.^[ref]

A useful independent test of the constant-coefficient model avoids numerical derivatives of noisy head readings. Its exact square root relation predictsFit a straight line to $\sqrt{h_i}$ against time over a chosen positive-head interval; the negative slope determines $b$. Repeat from the same initial head and compare slopes, then vary the head interval to test constancy of $C_d$. This is a proposed diagnostic derived from the draining tank with constant cross-sectional area, not an additional procedure or measured fit attributed to Price. Startup transients, uncertain outlet level, a changing jet shape and terminal low-head data can spoil the straight line. Independent collection of discharged volume provides a check on the tank-scale readout.

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 problemThe downward speed of the free surface is $Q/A$, smaller than the ideal jet speed by the factor $C_{d}a/A$. 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 yieldsThe 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, stipulateThe initial volume above outlet level is $AH=0.0250{\mathrm{m}}^3=25.0\mathrm{L}$. The initial ideal jet speed is $2.215\mathrm{m}{\mathrm{s}}^{-1}$, the initial volumetric flow rate is $Q_0=0.1373\mathrm{L}{\mathrm{s}}^{-1}$, 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 $Q_0/2=0.06866\mathrm{L}{\mathrm{s}}^{-1}$. Half the initial volume has already left at $t=T(1-1/\sqrt{2})=106.7\mathrm{s}$.

An independent mass conservation check integrates the outgoing volumetric flow rate:Multiply both sides by mass density to obtain discharged mass. A dimensional analysis check gives $[k]={\mathrm{m}}^{1/2}{\mathrm{s}}^{-1}$, $[Q]={\mathrm{m}}^3{\mathrm{s}}^{-1}$ 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.

Assume a sharp interface, uniformly mixed warm upper air with fixed reduced gravity $g^{\prime }=g\Delta \rho /{\rho }_0$, cool ambient air entering below, no wind, no wall heat exchange, negligible source volume, and quasi-steady Boussinesq flow through the small vents. Neglect the interior approach velocity. The hydrostatic buoyancy head between the two vents is ${\rho }_0{g}^{\prime }h$.

Both openings carry the same volume flux $q_v$. The discharge coefficient law gives a pressure loss ${\rho }_0{q}_v^2/(2C_d^2{A}^2)$ at each opening. Adding the two losses gives $q_v=C_{d}A\sqrt{g^{\prime }h}$, not the single-opening value with an extra factor $\sqrt{2}$. Cool replacement air displaces the warm layer, so $S\dot{h}=-q_v$. With $h(0)=H$, integrate to obtainThe ideal model sets $h=0$ thereafter. The formula applies only while the parenthesis is nonnegative; squaring a negative value would invent renewed filling. A finite aperture or a diffuse final interface invalidates the idealization near exhaustion.

Assume both openings have positive area, no imposed wind pressure, a light buoyant plume, and a Boussinesq approximation. Write $S=\pi Q_s$ and $B_s=\pi F_s$ for physical source volume and buoyancy flux. Include discharge coefficients $C_H,C_0$; if ideal openings are intended, set these to one. In a steady two-layer displacement ventilation state with interface height $z_{\ast }$, let the floor inflow be $q_0$ and roof outflow $q_H$. Volume and buoyancy balances giveThe hydrostatic pressure head across the warm upper layer must equal the two opening losses:A positive unprocessed lower layer needs $q_0>0$. At the boundary where this layer disappears, $z_{\ast }=0$, the plume can entrain no lower ambient fluid, so $q_0=0$, $q_H=S$ and $g_u^{\prime }=B_s/S$. The critical condition is therefore $S^3=2C_H^2{A}_H^2{B}_{s}H$.

To determine the inequality, consider a fully processed room. Its steady reduced gravity is $g^{\prime }=B_s/S$. If $p_0$ is the interior-to-exterior pressure difference at the floor, source-volume blocking of fresh floor inflow requires $p_0\ge 0$. Both openings then exhaust, withThe right-hand side is increasing in $p_0$ and has its minimum at zero. Consequently a fully processed state exists exactly whenThis is source-volume blocking of displacement ventilation. At equality floor inflow vanishes; above it the floor can also exhaust and no fresh lower-layer supply is needed. Below it, the floor pressure must be negative, admitting fresh fluid and supporting a positive lower layer. In fact the steady interface equation has a unique positive root: its loss side increases with $z_{\ast }$ because $Q(z_{\ast })$ increases, while its buoyancy-head side decreases.

The floor-opening area affects the positive-interface state and its approach, but drops out of the blocking threshold because floor flux is zero at the threshold. Equal splitting of pressure head, or using a source-free effective opening area at this transition, would miss this distinction. If the floor opening is sealed, the fresh-inflow argument no longer applies: a finite source with only roof exhaust instead cycles the room asymptotically as in the ceiling-opening case of part (c).

Define the indoor-to-outdoor reduced gravityBecause the whole opening lies above the interface, its indoor side contains upper-layer fluid of density ${\rho }_H$. Under hydrostatic pressure, the pressure difference varies linearly about a neutral pressure level. Equal opening geometry and equal discharge coefficients for inflow and outflow put that level at the vent midpoint. At vertical distance $y$ from it, the ideal Bernoulli equation gives speed $\sqrt{2g_V^{\prime }|y|}$.

Writing $Q_V$ for either one-way volumetric flow rate, integration over one half of the opening givesThe total unsigned exchange is $2Q_V$. The ideal sharp-edged inviscid model has $C_d=1$; an empirical $C_d<1$ represents contraction and losses.

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.