---
title: "Emptying time of cylinder-cone vessel"
subject: "Fluid Mechanics"
chapter: 6
key: "semester_3/fluid_mechanics:cylinder-cone-vessel-emptying-time"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/cylinder-cone-vessel-emptying-time
site: BCT Engineering
---

# Emptying time of cylinder-cone vessel

Fluid Mechanics · Chapter 6: Flow measurement

- Asked in: 2076 Asoj, 6 marks
- Video: [Bernoulli's Water Tank | How Long Will It Take to Drain?](https://www.youtube.com/watch?v=voDq_CjzRGE) — INTEGRAL PHYSICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/cylinder-cone-vessel-emptying-time
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-6

## Question

Find the time of emptying of cylindrical vessel attached with conical vessel as shown in the figure below. Orifice of diameter 10 cm is at the bottom of the tank. Take discharge coefficient as 0.6. There is no inflow into the tank. [Figure: cylindrical top portion 2 m diameter and 2 m high on top of a conical frustum 4 m high tapering to a 1 m diameter at the bottom where the orifice is]

## Short answer

Emptying time integrates -A(h)dh = Cd a sqrt(2g) sqrt(h) dt. Cylindrical part (6 m to 4 m): t1=135.3 s; conical frustum (4 m to 0): t2=280.9 s; total t≈416 s (6.9 min).

## Answer

**Given (assumed from the figure):** cylinder 2 m diameter, 2 m high on top of a conical frustum 4 m high tapering from 2 m diameter (top) to 1 m diameter (bottom); orifice $d = 0.10$ m at the bottom; $C_d = 0.6$. Initially full, so the head above the orifice is $H = 4 + 2 = 6$ m.

```text
   |<--2 m-->|
   |         |  2 m
   |_________|
    \       /
     \     /  4 m (frustum)
      \___/
       o  <- orifice 10 cm (bottom dia 1 m)
```

Orifice area $a = \dfrac{\pi}{4}(0.1)^2 = 0.007854$ m².

$$
k = C_d\,a\sqrt{2g} = 0.6\times 0.007854\times 4.4294 = 0.020873\ \text{m}^{5/2}/\text{s}
$$

Rate equation: $-A(h)\,dh = k\sqrt{h}\,dt\;\Rightarrow\; t = \dfrac{1}{k}\displaystyle\int\dfrac{A(h)}{\sqrt h}dh$.

#### Time to lower 6 m to 4 m (cylinder, $A = \pi$ m²)

$$
t_1 = \frac{A}{k}\cdot 2\left(\sqrt{6} - \sqrt{4}\right) = \frac{3.1416\times 2\times 0.4495}{0.020873} = 135.3\ \text{s}
$$

#### Time to lower 4 m to 0 (frustum)

At height $h$ above the orifice, the diameter is $D = 1 + h/4$ (1 m at $h = 0$, 2 m at $h = 4$), so

$$
A(h) = \frac{\pi}{4}\left(1 + \frac{h}{4}\right)^2 = \frac{\pi}{4}\left(1 + 0.5h + 0.0625h^2\right)
$$

$$
\begin{aligned}
t_2 &= \frac{\pi}{4k}\int_0^4\left(h^{-1/2} + 0.5h^{1/2} + 0.0625h^{3/2}\right)dh\\
&= \frac{\pi}{4k}\left[2h^{1/2} + \frac{1}{3}h^{3/2} + 0.025h^{5/2}\right]_0^4\\
&= \frac{\pi}{4k}\left[4 + 2.667 + 0.8\right] = \frac{0.7854\times 7.467}{0.020873} = 280.9\ \text{s}
\end{aligned}
$$

#### Total time

$$
t = t_1 + t_2 = 135.3 + 280.9 = 416.3\ \text{s} \approx 6.94\ \text{min}
$$

**Answer: time of emptying $\approx 416$ s (about 6 min 56 s).**

## Check yourself

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

1. Which relation gives emptying time of a vessel through a bottom orifice?
   A. t = A*H/(Cd*a)
   B. t = k*H^2/A
   C. t = (1/k) * integral of A(h)/sqrt(h) dh, with k = Cd*a*sqrt(2g)
   D. t = (1/k) * integral of A(h)*sqrt(h) dh
   Answer: C. From -A(h)dh = Cd*a*sqrt(2gh)dt, so t = (1/k) integral A(h)/sqrt(h) dh. (Key point: Rate equation)

2. Why is the calculation done in two parts?
   A. Flow stops at 4 m
   B. The orifice changes size
   C. The coefficient changes at 4 m
   D. The cross-sectional area is constant in the cylinder but varies with h in the frustum
   Answer: D. Cylinder area is constant (pi m^2); frustum area varies as (pi/4)(1 + h/4)^2, so each is integrated separately. (Key point: Splitting into stages)

3. At height h above the orifice in the frustum, what is the diameter?
   A. 2 - h/4 m
   B. 1 + h/4 m
   C. 1 + h/2 m
   D. 1 + 4h m
   Answer: B. Diameter grows from 1 m at h = 0 to 2 m at h = 4 m, so D = 1 + h/4. (Key point: Frustum diameter)

4. What is the total time of emptying?
   A. About 696 s
   B. About 135 s
   C. About 281 s
   D. About 416 s
   Answer: D. t1 = 135.3 s (cylinder) plus t2 = 280.9 s (frustum) gives about 416 s. (Key point: Final result)

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