Engineering calculator reviewed for preliminary design use · Last updated: March 2026
Calculate tank draining time using Torricelli's theorem from tank cross-section area, orifice area, and liquid head. Apply to batch operation scheduling, tank cleaning, and emergency drain planning.
What this calculator is used for
Calculates the time required to drain a tank through a bottom or side orifice/nozzle. Based on Torricelli's theorem (v = √(2gh)), it accounts for discharge coefficient, orifice area, and initial/final liquid level.
Typical engineering use cases
Estimating drain time for tank cleaning and inspection
Scheduling batch discharge and drain operations
Approximating emergency blowdown/drain duration
Supporting fire-water tank and sump drainage planning
Governing equation and methodology
Because outflow velocity falls as the level drops, Torricelli's theorem is integrated over time:
where A_tank is the tank cross-sectional area [m²], A_hole the orifice area [m²], Cd the discharge coefficient (≈0.61 sharp-edged, ≈0.82 rounded), h₁/h₂ the initial/final levels [m], and g gravitational acceleration. For complete draining set h₂ = 0.
Engineering assumptions and limitations
Assumes atmospheric venting and quasi-steady flow (slow level fall)
Constant tank cross-section (e.g. vertical cylinder)
Outlet piping friction is simplified to orifice flow
Ignores high viscosity, gas entrainment, and vortexing
Practical design notes
Drain time is inversely proportional to orifice area and depends on the square root of level, so the final portion of a full tank takes disproportionately long. When real outlet piping friction dominates, combine this with equivalent K-factors or a pressure-loss calculation. Viscous fluids lower Cd, and vortexing reduces effective area—add margin.
Worked Example
Given:
Vertical tank ID = 2.0 m (cross-section 3.14 m²)
Outlet orifice d = 50 mm (A₀ = 1.96×10⁻³ m²), Cd = 0.62
Initial liquid height h₁ = 3.0 m, draining to empty
Method: t = (A_tank/(Cd·A₀))·√(2/g)·(√h₁ − √h₂) = (3.14/(0.62·1.96×10⁻³))·√(2/9.81)·(√3 − 0).
Result: Drain time t ≈ 2020 s ≈ 33.7 min.
Interpretation: Outflow follows √h, so the last metre takes disproportionately long — most of the 34 min is spent draining the lower head, which matters for turnaround scheduling. If the outlet pipe is long, its friction (not Torricelli) will dominate and this gravity estimate becomes optimistic.
Common Mistakes & Misuse
Using the ideal Torricelli velocity √(2gh) without a discharge coefficient (~0.6–0.65), which over-predicts the drain rate by ~40%.
Assuming a constant cross-section when the tank is conical or horizontal-cylindrical, so the level-vs-time profile is wrong.
Forgetting that a sealed (un-vented) tank pulls a vacuum as it drains, choking the flow well below the open-vent prediction.
Applying inviscid theory to a viscous fluid or a long outlet pipe, where friction, not just the orifice, governs the drain time.
Frequently Asked Questions
What assumptions does Torricelli's theorem make?
Torricelli's theorem assumes inviscid, incompressible fluid with a free surface. It also assumes the tank cross-section is much larger than the outlet orifice. A discharge coefficient (0.6-0.65) corrects for real fluid effects.
Why does draining slow down as the tank empties?
The driving force is the hydrostatic head (liquid height above the outlet). As the liquid level drops, the head decreases, the exit velocity decreases, and the flow rate decreases — causing the draining to slow exponentially.
For preliminary estimation and educational use only. Results may depend strongly on assumptions, input data, fluid or material properties, and the range of validity of the underlying equation. Verify critical calculations independently and follow the applicable code, specification, and formal engineering review process before using any result for design, procurement, fabrication, operation, or safety decisions.