Tank Drain Time (Falling-Head Orifice)
How long a tank takes to empty through an orifice, integrating the falling head.
Example
You enter
- Tank cross-section area (ft²) 100
- Orifice diameter (in) 6
- Discharge coefficient Cd 0.6
- Starting head above orifice (ft) 9
You get
- Drain time 634.64
- T (min) 10.577
Details, formula, and sources
t = 2 A_t (sqrt(h1) - sqrt(h2)) / (Cd A_o sqrt(2 g)) for a constant-cross-section tank. A 100 ft^2 tank draining a 6 in orifice empties from 9 ft in 10.6 min, but reaches the last foot in only 7.1 - the flow slows as sqrt(h), so the final foot alone takes another 3.5 minutes. Prismatic tank, free discharge, steady Cd (~0.6). A design aid; the engineer of record governs.
A_o = pi/4 (d/12)^2; t = 2 A_t (sqrt(h1) - sqrt(h2)) / (Cd A_o sqrt(2 g)), g = 32.2 ft/s^2.
The falling-head (Torricelli) tank drain-time integral of the orifice equation, a standard hydraulics result, by name.
The falling-head drain-time integral is a public closed-form hydraulics result; the discharge coefficient is in the standard references.
Estimate. AHJ and licensed professional govern.
Field names used by the API: tank_area_ft2, d_in, cd, h1_ft, t_s, t_min
- Falling-head integral t = 2 A_t (sqrt(h1) - sqrt(h2)) / (Cd A_o sqrt(2 g)); the sqrt(h) head makes the final drawdown the slowestTorricelli / orifice hydraulics
- Prismatic tank constant cross-section area A_t with depth; a varying cross-section needs its own integrationscope of this tile