---
title: "Duration of flow after pump stops"
subject: "Hydraulics"
chapter: 4
key: "semester_4/hydraulics:pump-stop-duration-of-forward-flow"
source: https://www.bctengineer.com/q/semester_4/hydraulics/pump-stop-duration-of-forward-flow
site: BCT Engineering
---

# Duration of flow after pump stops

Hydraulics · Chapter 4: Unsteady flow in pipes

- Asked in: 2075 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/pump-stop-duration-of-forward-flow
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-4

## Question

A pump draws water from a reservoir and delivers it through a pipe 150 mm diameter, 90 m long, to a tank in which the free surface level is 8 m higher than that in the reservoir. The flow rate is steady at 0.05 m³/s until a power failure causes the pump to stop. Neglecting minor losses in the pipe and in the pump and assuming that the pump stops instantaneously, determine for how long flow into the tank continues after the power failure. The friction factor f may be taken as constant at 0.028 and elastic effects in the water or pipe material may be disregarded.

## Short answer

With rigid-column theory, the deceleration equation is integrated from initial flow to zero. The flow into the tank continues for 2.62 s after power failure.

## Answer

**Data.** $D = 0.15$ m, $L = 90$ m, static lift $H_s = 8$ m, $f = 0.028$, steady $Q_0 = 0.05$ m³/s. The pump stops instantaneously; minor losses and elasticity are neglected, so the water column decelerates as a rigid body under the static head (back-pressure) and friction.

#### Constants
$$
A = \frac{\pi}{4}(0.15)^2 = 0.01767\ \text{m}^2,\qquad I = \frac{L}{A} = \frac{90}{0.01767} = 5093.0\ \text{m}^{-1}
$$
$$
r = \frac{8fL}{\pi^2gD^5} = \frac{8(0.028)(90)}{\pi^2(9.81)(0.15)^5} = 2742.0\ \text{s}^2/\text{m}^5 \quad(\text{steady friction head } rQ_0^2 = 6.85\ \text{m})
$$

#### Equation of motion
$$
\frac{L}{gA}\frac{dQ}{dt} = -\left(H_s + rQ^2\right) \;\Rightarrow\; dt = -\frac{I}{g}\frac{dQ}{H_s + rQ^2}
$$
The flow into the tank stops when $Q = 0$. Integrating from $Q_0$ to 0:
$$
t = \frac{I}{g}\int_0^{Q_0}\frac{dQ}{H_s + rQ^2} = \frac{I}{g\sqrt{H_sr}}\tan^{-1}\left(Q_0\sqrt{\frac{r}{H_s}}\right)
$$
$$
Q_0\sqrt{\frac{r}{H_s}} = 0.05\sqrt{\frac{2742.0}{8}} = 0.9257,\qquad \tan^{-1}(0.9257) = 0.7468\ \text{rad}
$$
$$
t = \frac{5093.0}{9.81\sqrt{8(2742.0)}}(0.7468) = \frac{0.7468}{0.28528} = 2.62\ \text{s}
$$

**Answer:** The flow into the tank continues for about $2.62$ s (about 2.6 s) after the power failure.

## Check yourself

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

1. Why can the water column be treated as a decelerating rigid body here?
   A. Because elasticity of water and pipe is disregarded
   B. Because the pipe is short
   C. Because friction is zero
   D. Because the tank is empty
   Answer: A. With elastic effects neglected, the whole column moves together and obeys Newton's law with inertia L/(gA). (Key point: Rigid-column assumption)

2. Which equation governs the deceleration after pump stop?
   A. (L/(gA)) dQ/dt = -(H_s + rQ^2)
   B. (L/(gA)) dQ/dt = H_s - rQ^2
   C. dQ/dt = -rQ
   D. (L/(gA)) dQ/dt = rQ^2 - H_s
   Answer: A. Static head and friction both oppose forward flow, so both terms are negative. (Key point: Equation of motion)

3. What is the time for forward flow to stop?
   A. About 0.26 s
   B. About 2.62 s
   C. About 26 s
   D. About 8 s
   Answer: B. t = I tan^-1(Q0 sqrt(r/H_s)) / (g sqrt(H_s r)) = 0.7468/0.28528 = 2.62 s. (Key point: Time for flow to stop)

## Related in Hydraulics

- [Flow rate 2 s after pump failure](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Apump-failure-flow-rate-after-2s) — Hydraulics, chapter 4
- [Siphon with 90 mm pipe, number and length](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Asiphon-90mm-pipe-total-length) — Hydraulics, chapter 2
- [Pump power and pipe length with Colebrook](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Apump-power-pipe-length-colebrook) — Hydraulics, chapter 2

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