---
title: "Rapid closure time and peak hammer pressure"
subject: "Hydraulics"
chapter: 4
key: "semester_4/hydraulics:cast-iron-pipe-rapid-closure-time-and-length"
source: https://www.bctengineer.com/q/semester_4/hydraulics/cast-iron-pipe-rapid-closure-time-and-length
site: BCT Engineering
---

# Rapid closure time and peak hammer pressure

Hydraulics · Chapter 4: Unsteady flow in pipes

- Asked in: 2075 Baisakh, 2.5+2.5+3 marks
- Video: [NCH 3.3 Numerical on unsteady flow in pipe ( Minimum time of safe closure determination )](https://www.youtube.com/watch?v=Tzyw7UMpasI) — Aaradhya Elite Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/cast-iron-pipe-rapid-closure-time-and-length
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-4

## Question

A cast iron pipe of 300 mm diameter and 8 mm thick is 1500 m long. The pipe is to convey 200 litres per sec of water. a) Estimate the maximum time of closure of a valve at the downstream end that would be recognized as rapid closure? b) What is the peak water hammer pressure produced by rapid closure? c) What is the length of the pipe subjected to peak water hammer pressure if the time of closure is 2.0 sec? (For water E = 2200 MPa; for cast iron $E = 80\times10^9$ Pa)

## Short answer

Wave speed 1041 m/s; closure is rapid if ≤2.88 s. Peak pressure rise is 2.94 MPa. With 2 s closure time, peak pressure covers 459 m of pipe adjacent to valve.

## Answer

**Data.** $D = 0.3$ m, $t = 8$ mm, $L = 1500$ m, $Q = 200$ l/s, $K = 2200$ MPa $= 2.2\times10^9$ Pa, $E = 80\times10^9$ Pa for cast iron.

#### Velocity and wave speed
$$
V = \frac{Q}{\pi D^2/4} = \frac{0.2}{0.070686} = 2.829\ \text{m/s}
$$
$$
c = \frac{\sqrt{K/\rho}}{\sqrt{1 + \dfrac{KD}{Et}}} = \frac{\sqrt{2.2\times10^6}}{\sqrt{1 + \dfrac{(2.2\times10^9)(0.3)}{(80\times10^9)(0.008)}}} = \frac{1483.2}{\sqrt{1 + 1.0312}} = 1040.7\ \text{m/s}
$$

#### (a) Maximum time of closure for rapid closure
Closure is rapid if $t_c \le \dfrac{2L}{c}$ (the time for the pressure wave to travel to the reservoir and back):
$$
t_c \le \frac{2(1500)}{1040.7} = 2.883\ \text{s}
$$

#### (b) Peak water hammer pressure (rapid closure)
$$
\Delta p = \rho cV = 1000(1040.7)(2.829) = 2.945\ \text{MPa}\quad(\Delta H = \frac{cV}{g} = 300.2\ \text{m of water})
$$

#### (c) Length of pipe subjected to peak pressure for $t_c = 2.0$ s
Since $2.0\ \text{s} < 2L/c = 2.88$ s, the closure is rapid, but the closing takes a finite time, so only part of the pipe sees the full peak pressure. At a point $d$ from the valve, the last pressure increment of the closure arrives at $t = t_c + d/c$, while the negative wave reflected from the reservoir arrives at $t = (2L - d)/c$. The full peak $\rho cV$ is reached only if the last increment arrives before the reflection:
$$
t_c + \frac{d}{c} \le \frac{2L - d}{c} \;\Rightarrow\; d \le L - \frac{c\,t_c}{2}
$$
$$
L_{peak} = 1500 - \frac{1040.7(2.0)}{2} = 459.3\ \text{m}
$$
This is the length of pipe measured from the valve towards the reservoir that experiences the peak pressure; the remaining $c\,t_c/2 = 1040.7$ m next to the reservoir does not.

**Answer:** (a) $t_c \le 2.88$ s; (b) $\Delta p = 2.94$ MPa; (c) about $459$ m of the pipe adjacent to the valve experiences the peak pressure.

## Check yourself

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

1. When is valve closure called rapid?
   A. When closure time t_c is at most 2L/c
   B. When t_c exceeds 4L/c
   C. When t_c is less than L/2c only
   D. When velocity is under 3 m/s
   Answer: A. If the valve is closed before the reflected wave from the reservoir returns (2L/c), the full pressure develops. (Key point: Rapid closure criterion)

2. With L = 1500 m and c = 1040.7 m/s, what is the maximum time of rapid closure?
   A. 2.88 s
   B. 1.44 s
   C. 5.77 s
   D. 0.72 s
   Answer: A. 2L/c = 3000/1040.7 = 2.883 s. (Key point: Maximum rapid closure time)

3. What is the peak water hammer pressure (V = 2.829 m/s)?
   A. 2.94 MPa
   B. 0.29 MPa
   C. 29.4 MPa
   D. 1.47 MPa
   Answer: A. delta p = rho c V = 1000 x 1040.7 x 2.829 = 2.945 MPa (about 300 m of water). (Key point: Peak pressure)

4. For t_c = 2.0 s, what length of pipe next to the valve sees the peak pressure?
   A. L - c t_c / 2 = about 459 m
   B. c t_c / 2 = about 1041 m
   C. the whole 1500 m
   D. 2L - c t_c = about 919 m
   Answer: A. Full peak occurs only where the last increment arrives before the reservoir reflection: d <= L - c t_c/2 = 459.3 m. (Key point: Length subject to peak pressure)

## Related in Hydraulics

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