---
title: "Time to discharge pressurised tank"
subject: "Fluid Mechanics"
chapter: 6
key: "semester_3/fluid_mechanics:pressurised-tank-emptying-orifice"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/pressurised-tank-emptying-orifice
site: BCT Engineering
---

# Time to discharge pressurised tank

Fluid Mechanics · Chapter 6: Flow measurement

- Asked in: 2078 Bhadra, 6+2 marks
- Video: [Find Time Taken to Empty a Tank - Continuity Equation Example Problem](https://www.youtube.com/watch?v=HQ366gVl90w) — Brian Bernard
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/pressurised-tank-emptying-orifice
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-6

## Question

A pressurised 2 m diameter tank of water has a 10 cm dia orifice at the bottom, where water discharges to the atmosphere. The water level initially is 3 m above the outlet. The tank air pressure above water level is maintained at 450 kPa absolute and the atmospheric pressure is 100 kPa. Neglecting the frictional effects, determine (i) how long it will take for half of the water in the tank to discharge and (ii) the water level in the tank after 10 sec.

## Short answer

Effctive head is h + 35.678 m; using unsteady Bernoullli, time to drop from 3 m to 1.5 m is t = (2A/(a√(2g)))[√(3+35.678) - √(1.5+35.678)] ≈ 22.0 s. After 10 s, solving √(h+35.678) = 6.1638 gives h ≈ 2.31 m.

## Answer

**Given:** tank $D = 2$ m, so $A = \dfrac{\pi}{4}(2)^2 = 3.1416$ m²; orifice $d = 0.1$ m, $a = \dfrac{\pi}{4}(0.1)^2 = 0.007854$ m²; $h_0 = 3$ m; air pressure 450 kPa absolute, atmospheric 100 kPa; $C_d = 1$ (no friction); air pressure kept constant.

**Pressure head of air above the water (gauge):**

$$
h_p = \frac{450 - 100}{9.81} = 35.678\ \text{m}
$$

The effective head at the orifice is $h + h_p$, so the exit velocity is $V = \sqrt{2g(h + h_p)}$.

**Rate equation:** $-A\dfrac{dh}{dt} = a\sqrt{2g(h + h_p)}$.

$$
t = \frac{A}{a\sqrt{2g}}\int_{h_2}^{h_1}\frac{dh}{\sqrt{h + h_p}} = \frac{2A}{a\sqrt{2g}}\left[\sqrt{h_1 + h_p} - \sqrt{h_2 + h_p}\right]
$$

$$
\frac{a\sqrt{2g}}{2A} = \frac{0.007854\times 4.4294}{2\times 3.1416} = 0.0055368\ \text{m}^{1/2}/\text{s}
$$

#### (i) Time for half the water to discharge

The water level falls from 3 m to 1.5 m (half the volume, since the tank is cylindrical).

$$
\begin{aligned}
\sqrt{3 + 35.678} &= 6.2192,\qquad \sqrt{1.5 + 35.678} = 6.0974\\
t &= \frac{6.2192 - 6.0974}{0.0055368} = 22.0\ \text{s}
\end{aligned}
$$

#### (ii) Level after 10 s

$$
\begin{aligned}
\sqrt{h + h_p} &= 6.2192 - 0.0055368\times 10 = 6.1638\\
h + h_p &= 37.992\\
h &= 37.992 - 35.678 = 2.31\ \text{m}
\end{aligned}
$$

**Answer: (i) $t \approx 22.0$ s; (ii) water level after 10 s $\approx 2.31$ m above the outlet.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Gauge pressure head of air?
   A. 45.9 m
   B. 3 m
   C. 35.68 m
   D. 350 m
   Answer: C. 350 kPa / 9.81. (Key point: Pressure head)

2. Rate equation?
   A. -A dh/dt = a sqrt(2gh)
   B. A dh = a dt
   C. dh/dt = const
   D. -A dh/dt = a sqrt(2g(h+hp))
   Answer: D. Effective head includes air pressure head. (Key point: Rate equation)

3. Level when half discharged?
   A. 1.5 m
   B. 3 m
   C. 0 m
   D. 1 m
   Answer: A. Cylindrical tank. (Key point: Half water)

4. Time for half?
   A. 10 s
   B. 22.0 s
   C. 60 s
   D. 2.2 s
   Answer: B. (6.2192-6.0974)/0.0055368. (Key point: Time)

## Related in Fluid Mechanics

- [Time to lower level in two-orifice tank](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Atwo-orifice-tank-lowering-time) — Fluid Mechanics, chapter 6
- [Absolute and gauge pressure, mercury manometer](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Aabsolute-gauge-pressure-vapour-mercury) — Fluid Mechanics, chapter 2
- [Rayleigh's method for orifice discharge](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Aorifice-discharge-rayleigh-method) — Fluid Mechanics, chapter 10
- [Gauge pressure in cleaning fluid tank](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Acleaning-fluid-tank-gauge-pressure) — Fluid Mechanics, chapter 2
- [Water mass for pressurised vessel](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Apressure-vessel-bulk-modulus) — Fluid Mechanics, chapter 1
- [Pressure diagram in rotating closed tank](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Aclosed-rotating-tank-pressure-diagram) — Fluid Mechanics, chapter 2

## Related in other subjects

- [Pipe outlet minimum water level, 50 lps](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Apipe-outlet-min-water-level-50lps) — Irrigation and Drainage Engineering, chapter 7
- [Three reservoirs with pressurised closed tank](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Athree-reservoir-closed-tank-pressure) — Hydraulics, chapter 3
- [Volumetric efficiency from orifice air measurement](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Avolumetric-efficiency-orifice-air-consumption) — Internal Combustion Engines, chapter 8
- [Storage tank capacity and water level, 20000 L/day](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Areservoir-level-river-20000) — Water Supply Engineering, chapter 7, asked 2×