Tank Drain & Torricelli Decay

tools.tank-drain.intro

Worked example — Tank Drain & Torricelli Decay

Tank cross-section (A) (m²)	0.0707
Hole area (a) (m²)	7.85 × 10⁻⁵
Time to Empty (T) (s)	655.4
Current Water Level (h) (m)	1.000
Current Flow Rate (Q) (m³/s)	0.0002
Time to 50% Level	29.3%
Bottom / Top Time Ratio	2.414
Initial Water Level (H)	1.00 m

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

A draining tank has no half-life 🖖

Everyone expects this to decay like a charged capacitor, and it does not. Outflow follows √h rather than h, so the level falls as H(1 − t/T)² — a parabola that reaches zero at a finite time. Match an exponential to the same half-emptying moment and at the instant this tank is actually dry it still claims 9.4% is left, and it never empties at all. That is the difference between a rate set by the amount and a rate set by its square root.

Four times the depth is twice the wait 🖖

T = (A / C_dA₀)√(2H/g), and H is under a square root while both areas are not. Fill the same tank four times deeper and it takes twice as long, not four times. Halve the hole and it takes exactly four times as long, because area goes with the square of the diameter. The depth is the forgiving term here and the hole is the brutal one.

The hole is smaller than the hole you drilled 🖖

The discharge coefficient is not a fudge factor. Water approaching a sharp-edged orifice cannot turn the corner instantly, so the stream keeps narrowing after it has left the wall and reaches its true minimum a little outside — the vena contracta. A sharp hole runs near 0.62, so the effective opening is under two thirds of the one you made. Set C_d to 1 and the tank empties in 62% of the real time, which is the size of the error you make by trusting the drawing.

Problem solved in full

Why the bottom half of a barrel takes 2.4 times as long as the top 8 steps
A 300 mm barrel is filled 1.00 m deep and a 10 mm hole is opened in the bottom. How long does it take to empty, and how much of that time goes on the last half of the water?
1. Two areas carry the whole problem. The barrel's cross-section says what a millimetre of level is worth in litres; the hole's says how fast those litres can get out.
  
  $A = \frac{π d ^{2}}{4} = \frac{π ( 0.30 ) ^{2}}{4} = 0.0707 m^{2}, a = \frac{π ( 0.010 ) ^{2}}{4} = 7.85 \times 1 0^{- 5} m^{2}$
2. The speed at the hole is Torricelli's again, set by the depth still standing above it. The discharge coefficient is there because the jet pinches to about 62% of the hole as it leaves. Multiply by the hole area for the volume per second, then divide by the barrel area for the speed the level drops.
  
  $\frac{d h}{d t} = - \frac{C _{d} a 2 g h}{A}$
3. The depth is on both sides, so separate it and integrate from full to empty. This is the only calculus in the problem, and the shape of everything below follows from it.
  
  $\int_{H}^{0} \frac{d h}{h} = - \frac{C _{d} a 2 g}{A} \int_{0}^{T} d t ⟹ 2 H = \frac{C _{d} a 2 g}{A} T$
4. Rearranged, that is the drain time. The numbers go in.
  
  $T = \frac{A}{C _{d} a} \frac{2 H}{g} = \frac{0.0707}{0.62 \times 7.85 \times 1 0 ^{- 5}} \frac{2 \times 1.00}{9.81} = 655.4 s$
5. Stopping the same integral at a depth h instead of at zero, and turning it round, gives the level at any moment. It is a parabola: it reaches zero at T and stays there. A capacitor never quite finishes discharging. A barrel does, because the last of the water is still being pushed by the last of the depth.
  
  $h (t) = H (1 - \frac{t}{T})^{2}$
6. Set the level to half and solve for the time. It is not half of T.
  
  $\frac{H}{2} = H (1 - \frac{t}{T})^{2} ⟹ \frac{t}{T} = 1 - \frac{1}{2} = 0.293$
7. The top half is gone in 29.3% of the run, which leaves the bottom half the other 70.7%. Nothing in that ratio knows the depth, the diameter, the gravity or the coefficient.
  
  $\frac{T - t _{1/2}}{t _{1/2}} = \frac{1/ 2}{1 - 1/ 2} = 1 + 2 = 2.414$
8. T carries a square root of H and nothing else about the water, so filling the same barrel to 2.00 m does not double the wait. It multiplies it by 1.414.
  
  $\frac{T ( 2.00 )}{T ( 1.00 )} = \frac{2.00}{1.00} = 1.414 ⟹ T = 926.9 s$
Answer

655.4 seconds, and 70.7% of them go on the bottom half. The level falls as H(1 − t/T)², a parabola that reaches empty at a finite time rather than an exponential that only approaches it. Half the water is gone after 29.3% of the run, which leaves the bottom half taking 1 + √2 = 2.414 times as long as the top. There is no tank anywhere in that number.

$T = \frac{A}{C _{d} a} \frac{2 H}{g} \propto H$

Learning path

Water out of a tank: one speed, spent two ways

References (1)

Torricelli's law applied to quasi-steady tank draining kinetics and parabolic depth decay: E. Torricelli, De motu gravium naturaliter descendentium. Florence, 1644.

By the 42Math team · Updated 2026-09-01

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

All Physics lessons →

Tank Drain & Torricelli Decay

A draining tank has no half-life 🖖

Four times the depth is twice the wait 🖖

The hole is smaller than the hole you drilled 🖖

Example problems