Understanding the tank drain time calculator — concept and principle
This calculator estimates how fast water leaves a hole at the bottom of an open, vertical cylindrical tank and how long the tank takes to empty. Enter the water height, tank diameter, hole diameter and discharge coefficient to get the current outflow speed, the time to drain completely and the starting flow rate. It is useful for draining rain barrels, pools and process tanks, and for classic physics and calculus problems.
How it is calculated
- Outflow speed (Torricelli's law): v = √(2gH), with g = 9.80665 m/s².
- Initial flow: Q₀ = C_d · a · √(2gH), where a = πd²/4 is the hole area.
- Time to empty: t = (A ÷ (C_d · a)) · √(2H ÷ g), where A = πD²/4 is the tank area; the result is divided by 60 for minutes.
The principle
Evangelista Torricelli showed that liquid leaves an opening at the speed a body would reach falling freely from the surface height: v = √(2gH). It follows from Bernoulli's equation between the free surface (atmospheric pressure, nearly still) and the jet. Because the jet contracts just after a sharp-edged hole (vena contracta) and loses a little energy, the real flow is the ideal flow times C_d, about 0.6 for a sharp-edged orifice. As the level falls, the speed falls with √H. Writing A·dH/dt = −C_d·a·√(2gH) and integrating from H to 0 gives the drain-time formula. A neat consequence: the tank takes exactly twice as long to empty as it would if the initial flow continued unchanged.
Worked example
- Inputs: H = 2 m, D = 1 m, d = 25 mm, C_d = 0.61.
- v = √(2 × 9.80665 × 2) = 6.263 m/s (shown as 6.263114241).
- Area ratio A/a = (1000/25)² = 1600; divided by C_d: 1600 ÷ 0.61 = 2622.95.
- √(2H/g) = √(4 ÷ 9.80665) = 0.63866 s; t = 2622.95 × 0.63866 = 1675.2 s = 27.92 min (shown as 27.91955906).
- Q₀ = 0.61 × 0.00049087 × 6.2631 = 0.0018754 m³/s = 1.875 L/s.
| Water height (m) | Outflow speed now (m/s) |
|---|---|
| 0.5 | 3.131557121 |
| 1 | 4.428690551 |
| 1.5 | 5.42401604 |
| 2 | 6.263114241 |
| 2.5 | 7.002374597 |
| 3 | 7.670717046 |
| 4 | 8.857381103 |
How to read the result
Drain time grows with √H but with the square of the tank-to-hole diameter ratio. Doubling the hole diameter cuts the time to a quarter; doubling the water height increases it only by about 41%. Most of the time is spent on the last, slow part: the top half of the height drains in about 29% of the total time.
Common mistakes
- Entering the hole diameter in metres instead of millimetres.
- Using a pipe or valve outlet with C_d = 0.61; outlets with short pipes and fittings behave differently.
- Applying it to tanks that are not vertical cylinders (cones, horizontal tanks), where A changes with height.
Frequently asked questions
Why is the outflow speed shown without C_d?
It is the ideal Torricelli speed. Actual jet speed is slightly lower (velocity coefficient about 0.97–0.99), and the contraction mainly reduces the effective area.
Does a sealed tank drain the same way?
No. Without a vent, pressure above the water drops and the flow slows or gurgles.
How do I include a hole above the bottom?
Measure H from the water surface down to the hole, not to the tank floor.
Sources: E. Torricelli, Opera geometrica (De motu gravium), 1644 · F. M. White, Fluid Mechanics, McGraw-Hill