Estimate how long a vertical cylindrical tank takes to drain through a bottom orifice. The tool combines Torricelli's theorem v=√(2gh) with a quasi-steady mass balance, so adjusting tank diameter, orifice diameter and head gives the full drain time and outlet velocity in real time — a quick sizing aid for sumps, chemical storage tanks and emergency dump systems.
Parameters
Tank diameter D
m
Internal diameter of the cylindrical tank (A_t = π(D/2)²)
As the level falls, the outlet velocity drops with √h and the jet from the orifice gets slower. The numbers below the tank show the current head and outlet velocity.
Outlet velocity v from Torricelli's theorem (= the speed a stone would reach falling from height h). The drain time scales with √h₀ and inversely with the orifice area: doubling the tank diameter multiplies it by 4, doubling the orifice diameter divides it by 4.
The head drops as the square of a quantity that decays linearly from √h₀. The result is a "concave-up" curve — the level falls fast at first and then slows as h approaches zero.
What is the Tank Drain Time Simulator?
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Professor, if I have a bucket with a small hole in the bottom, can I actually calculate how long it takes to empty?
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You can. In the 17th century Torricelli (Galileo's student) showed that the speed of the water leaving the hole is v = √(2gh) — exactly the speed a stone would reach if dropped from the water surface down to the hole. It is the energy-conservation answer, but at the time it was an astonishing result.
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Wait, so water comes out at the same speed as a free-falling rock? With h₀ = 2 m that gives about 6.3 m/s, and that matches the default value on the right.
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Yes — √(2·9.81·2) ≈ 6.26 m/s, governed entirely by gravity. Then you write a mass balance ("volume leaving per unit time = tank area times the rate the level falls") and integrate h, and the time to empty drops out in closed form: t = (A_t/(C_d·A_o))·√(2h₀/g). For the default tank that is about 1648 seconds, or roughly 27 and a half minutes.
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Interesting — but the "half-drain time" on the left shows only 483 seconds, about 8 minutes. So half the water comes out in less than a third of the total time. The second half is much slower.
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Great catch. That is the signature of Torricelli draining. Because the outlet velocity falls as √h, the flow rate drops as the head drops. The first 50% of volume needs about 29% of the time and the last 50% needs about 71%. Operators often complain that "it's almost empty but never quite finishes" — that is exactly why. In practice the last metre is often pumped out actively rather than waiting for gravity.
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What is this discharge coefficient C_d? It multiplies the hole area, but it looks a bit mysterious.
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As the flow passes through the hole, the streamlines contract just downstream of the orifice — this is the "vena contracta". For a sharp-edged hole in a plate, the contracted jet area is only about 0.62 of the geometric hole area, and there is some friction on top of that. C_d packages those losses into one factor. Typical values are 0.60–0.65 for a sharp orifice and 0.95+ for a smoothly rounded bell-mouth inlet. Try the slider: pushing C_d from 0.62 toward 0.85 shortens the drain time inversely.
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Is this calculation actually used in real engineering, or is it just a textbook exercise?
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It is used all the time. Emergency dump times for chemical storage tanks, release schedules for stormwater detention basins, drain plugs on water heaters and boilers, early-phase tank leak scenarios for nuclear plants, and even the dial markings on the old water clocks (clepsydra) come from this equation. As long as A_t/A_o is large (above ~100) and the exit is open to atmosphere, it gives accuracy good enough for first-pass design. When those assumptions break — pressurised tanks, tapered shapes, long piping downstream — you go back to the full Bernoulli equation with a loss term.
Frequently Asked Questions
For a vertical cylindrical tank of constant cross-section draining freely through a bottom orifice, Torricelli's theorem gives the outlet velocity v = √(2gh). Combining this with the tank mass balance (A_t·dh/dt = −C_d·A_o·√(2gh)) and integrating from the initial level h₀ to zero gives the closed-form drain time t_drain = (A_t / (C_d·A_o)) · √(2h₀/g), where A_t is the tank cross-section, A_o is the orifice area, C_d is the discharge coefficient (0.60–0.65 for a sharp-edged orifice) and g is gravity. This tool evaluates that closed-form solution directly.
Because the outlet velocity v = √(2gh) scales with the square root of head, so the flow slows down as the level drops. The time to drain to half height is t_half = (A_t/(C_d·A_o))·√(2/g)·(√h₀ − √(h₀/2)) ≈ 0.293·t_drain. In other words the first half of the drain takes only ~29% of the total time, while the second half takes ~71%. With the defaults (D=2m, d_o=50mm, h₀=2m, C_d=0.62) you get t_drain ≈ 1648 s and t_half ≈ 483 s, clearly showing that asymmetric behaviour.
C_d is the ratio of actual to ideal flow rate and is less than 1 because of the vena contracta (the streamlines contract just downstream of a sharp orifice) plus friction losses. Typical values are 0.60–0.65 for a sharp-edged orifice, 0.95–0.99 for a bell-mouth inlet, and 0.50–0.53 for a short cylindrical (Borda) tube. Use 0.62 as a starting estimate for sharp-edged orifices, then refine with bench tests or CFD. This tool lets you sweep C_d from 0.50 to 0.85 so you can see the sensitivity directly.
The closed-form solution rests on quasi-steady assumptions: (1) the free-surface velocity is much smaller than the jet velocity (A_t/A_o is large — 1600 for the default settings), (2) the free surface is open to atmosphere, (3) the orifice exit is also at atmospheric pressure with no back-pressure, (4) viscous losses are lumped into C_d, and (5) the tank cross-section is independent of height (cylindrical). For tapered tanks, pressurised tanks, siphon outflow or cases where a long pipeline follows the orifice, you need extra loss terms or a fully transient Bernoulli formulation.
Real-World Applications
Emergency dump design for process tanks: Chemical-storage and fuel tanks must be able to dump their contents to a safe sump within a prescribed time (often 10–15 minutes) in case of leak or fire. The Torricelli formula in this tool is the very first step in sizing the emergency drain valve — the minimum orifice area is back-solved from t_drain — before piping losses and pumped assistance are added in the detailed design. Because the second half drains slowly, the last metre is usually evacuated by an active pump rather than relying on gravity alone.
Stormwater detention basins and release planning: Detention basins that hold urban storm runoff are typically subject to maximum discharge limits to the downstream river or sewer. The Torricelli flow Q = C_d·A_o·√(2gh) is used as the design rule for the release-gate geometry so that the peak discharge at full pond stays under the regulatory limit. The same closed form gives a quick estimate of the time response when a gate is suddenly opened during a seismic event.
Hot-water tanks and boiler drain plugs: Domestic heat-pump tanks and commercial boilers have drain plugs for maintenance. Specifications like "must be empty within 30 minutes" are checked using this formula given the tank volume and plug bore. One caveat: if the drain hose ends below the tank bottom, a siphon effect adds to the effective head and the tank empties faster than the simple formula predicts.
Water clocks (clepsydra) and physics teaching: Ancient Egyptian and Greek water clocks scratched their hour marks empirically, long before Torricelli's theorem was known. To make the marks equally spaced you need a non-cylindrical tank, a shape problem that was finally solved analytically in the Galileo–Torricelli era. The simple cylinder case remains a classic teaching demo: with a stopwatch and a ruler students can reproduce the curve this tool generates.
Common Misconceptions and Pitfalls
The biggest pitfall is to set C_d = 1.0 in the calculation. Textbooks often write the ideal flow as Q = A_o·√(2gh), and plugging that directly into spreadsheets over-predicts the real flow by about 38% and under-predicts the drain time by the same margin. The classic "it took 1.6 times longer than predicted" failure on real equipment usually traces back to this. Always carry C_d ≈ 0.62 for a sharp orifice, and run the slider from 0.62 to 0.95 to see how much rounding or chamfering at the inlet could improve things.
The second trap is to use this formula when a long pipe follows the orifice. The tool assumes the orifice exit discharges straight to atmosphere. In practice the drain plug is often connected to several metres of pipework, and pipe friction, fittings and exit losses must be added. The proper form becomes v = √(2gh / (1 + Σf·L/D + ΣK)), which effectively pushes C_d down to 0.3–0.5 for long piping. The longer the downstream pipe, the more optimistic the bare Torricelli formula becomes.
The third trap is to treat the free-surface velocity as the same order as the jet velocity. The quasi-steady solution needs A_t/A_o to be large — about 100 or more (the default in this tool is 1600). For tanks where A_t/A_o is small (say a 20 cm tank with a 10 cm hole) the surface drops nearly as fast as the jet rises and the simple Torricelli formula loses accuracy. Outflow problems with small area ratios (dam-break events, nuclear-vessel breaches) lie outside the scope of this tool and need a fully transient Bernoulli/ODE model.