---
title: "Time to lower level in two-orifice tank"
subject: "Fluid Mechanics"
chapter: 6
key: "semester_3/fluid_mechanics:two-orifice-tank-lowering-time"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/two-orifice-tank-lowering-time
site: BCT Engineering
---

# Time to lower level in two-orifice tank

Fluid Mechanics · Chapter 6: Flow measurement

- Asked in: 2075 Chaitra, 8 marks
- Video: [Time Required to Emptying the Tank | Case - 1 | Unsteady Flow | Fluid Mechanics](https://www.youtube.com/watch?v=2vj2UvQq5z4) — The Content Paradox
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/two-orifice-tank-lowering-time
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-6

## Question

A tank of constant cross-sectional area of 3.2 m² has two orifices each 8.8 mm² in area in one of its vertical sides at heights 5 m and 2 m respectively above the bottom of the tank. Calculate the time taken to lower the water level from 8 m to 3 m above bottom of tank. Assume $C_d$ = 0.62.

## Short answer

Stage 1 (8 m to 5 m) both orifices active, stage 2 (5 m to 3 m) only lower orifice. With k=Cd a sqrt(2g)=2.417e-5 m^(5/2)/s, total time t≈3.21e5 s (≈89 h).

## Answer

**Given:** tank area $A = 3.2$ m²; two orifices each $a = 8.8$ mm² $= 8.8\times 10^{-6}$ m² at 5 m and 2 m above the bottom; $C_d = 0.62$; level falls from 8 m to 3 m.

The upper orifice (at 5 m) works only while the level is above 5 m. So the problem has two stages:

- Stage 1: level 8 m to 5 m, both orifices discharging.
- Stage 2: level 5 m to 3 m, only the lower orifice (at 2 m) discharging.

$$
k = C_d\,a\sqrt{2g} = 0.62\times 8.8\times 10^{-6}\times 4.4294 = 2.4167\times 10^{-5}\ \text{m}^{5/2}/\text{s}
$$

Head over the upper orifice $= h - 5$; over the lower orifice $= h - 2$ ($h$ = level above tank bottom).

#### Stage 1 (8 m to 5 m)

$$
-A\,dh = k\left[\sqrt{h - 5} + \sqrt{h - 2}\right]dt
$$

Rationalise: $\dfrac{1}{\sqrt{h-5} + \sqrt{h-2}} = \dfrac{\sqrt{h-2} - \sqrt{h-5}}{3}$.

$$
\begin{aligned}
t_1 &= \frac{A}{k}\int_5^8\frac{\sqrt{h-2} - \sqrt{h-5}}{3}dh = \frac{A}{k}\cdot\frac{2}{9}\left[(h-2)^{3/2} - (h-5)^{3/2}\right]_5^8\\
&= \frac{3.2}{2.4167\times 10^{-5}}\times\frac{2}{9}\,[\ldots]
\end{aligned}
$$

Evaluating the bracket: at $h = 8$: $6^{3/2} - 3^{3/2} = 14.697 - 5.196 = 9.501$; at $h = 5$: $3^{3/2} - 0 = 5.196$. Difference $= 4.305$.

$$
t_1 = \frac{3.2}{2.4167\times 10^{-5}}\times\frac{2}{9}\times 4.305 = 126\,663\ \text{s}
$$

#### Stage 2 (5 m to 3 m, lower orifice only)

$$
t_2 = \frac{2A}{k}\left[\sqrt{5-2} - \sqrt{3-2}\right] = \frac{2\times 3.2}{2.4167\times 10^{-5}}\times(1.7321 - 1) = 193\,864\ \text{s}
$$

#### Total

$$
t = t_1 + t_2 = 126\,663 + 193\,864 = 320\,527\ \text{s} \approx 89.0\ \text{h}
$$

**Answer: $t \approx 3.2\times 10^{5}$ s (about 89 hours) for the data as given.** The very long time results from the tiny orifice areas (8.8 mm²). If the areas were 8.8 cm², all times would be 100 times smaller (about 3205 s, i.e. 53 min).

## Check yourself

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

1. When does the upper orifice at 5 m stop discharging?
   A. When the level reaches 3 m
   B. When the level falls below 5 m above the bottom
   C. It never stops
   D. When the level falls below 8 m
   Answer: B. Head over the upper orifice is h - 5, so it works only while the level exceeds 5 m. This splits the problem into two stages. (Key point: Upper orifice stops)

2. In stage 1 (8 m to 5 m), what is the flow equation?
   A. -A dh = 2k*sqrt(h)*dt
   B. -A dh = k*sqrt(h - 5)*dt
   C. -A dh = k*[sqrt(h - 5) + sqrt(h - 2)]*dt
   D. -A dh = k*sqrt(h - 2)*dt
   Answer: C. Both orifices discharge with heads h - 5 and h - 2, so flows add. (Key point: Stage 1 equation)

3. How is 1/[sqrt(h-5) + sqrt(h-2)] integrated?
   A. Take the average of the two heads
   B. Expand as a binomial series
   C. Multiply by the conjugate; the difference is 3, giving [sqrt(h-2) - sqrt(h-5)]/3
   D. Substitute h = 0
   Answer: C. (h-2) - (h-5) = 3, so the conjugate gives a sum of easy power terms. (Key point: Rationalising)

4. In stage 2 (5 m to 3 m), which formula applies?
   A. t2 = (2A/k)[sqrt(5-2) - sqrt(3-2)]
   B. t2 = (A/2k)[sqrt(3) - sqrt(5)]
   C. t2 = (A/k)[sqrt(5) - sqrt(3)]
   D. t2 = (2A/k)[sqrt(5) - sqrt(3)]
   Answer: A. Only the lower orifice (centre at 2 m) acts, so t = (2A/k)(sqrt(H1) - sqrt(H2)) with heads 3 m and 1 m. (Key point: Stage 2 time)

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