---
title: "Hammer pressure for 4 s and 60 s closure"
subject: "Hydraulics"
chapter: 4
key: "semester_4/hydraulics:steel-pipeline-61cm-hammer-4s-60s"
source: https://www.bctengineer.com/q/semester_4/hydraulics/steel-pipeline-61cm-hammer-4s-60s
site: BCT Engineering
---

# Hammer pressure for 4 s and 60 s closure

Hydraulics · Chapter 4: Unsteady flow in pipes

- Asked in: 2071 Magh, 8 marks
- Video: [Water hammer: slow closure of the valve and the Michaud's equation](https://www.youtube.com/watch?v=oh7Q6P7UVI4) — Marcos Vianna
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/steel-pipeline-61cm-hammer-4s-60s
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-4

## Question

A steel pipeline ($\epsilon$ = 0.046 mm) 61 cm in diameter and 3.2 km long laid freely at its lower end under a head of 61 m. What water-hammer pressure would develop if a valve at the outlet were closed in 4 sec? 60 sec? Wall thickness = 0.5 cm for both cases of closure. Compute the stress that would develop in the walls of the pipe near the valve. If the working stress of steel is taken as 16,000 psi, what would be the minimum time of safe closure? Consider $E_{water} = 2.17\times10^9$ N/m² and $E_p = 1.9\times10^{11}$ N/m².

## Short answer

Initial velocity 4.30 m/s, wave speed 952 m/s, critical time 6.72 s. 4 s closure (rapid): hammer pressure 4.09 MPa, hoop stress 286 MPa. 60 s closure (gradual): 0.229 MPa, stress 50.5 MPa. With allowable stress 110.3 MPa, minimum safe closure time is 11.4 s.

## Answer

First, the velocity before closure is found from the head loss in the pipe. Then the closure time is compared with $2L/c$: if $t_c<2L/c$ the closure is rapid ($\Delta p=\rho cV$); otherwise it is gradual ($\Delta p=\rho LV/t_c$).

#### Assumptions
The pipe discharges freely at the outlet under a head $H=61$ m. Take $\nu=10^{-6}$ m²/s and $g=9.81$ m/s². Losses = friction plus the exit velocity head.

#### Step 1: Initial velocity
$\varepsilon/D=0.046\times10^{-3}/0.61=7.5\times10^{-5}$. Solve
$$
H=\left(1+f\frac{L}{D}\right)\frac{V^2}{2g}
$$
with the Colebrook equation for $f$ (iterated): $f=0.0121$ at $Re\approx2.62\times10^6$.
$$
V=\sqrt{\frac{2\times9.81\times61}{1+0.0121\times3200/0.61}}=4.30\ \text{m/s}
$$

#### Step 2: Wave celerity
$$
c=\sqrt{\frac{K/\rho}{1+KD/(Et)}}=\sqrt{\frac{2.17\times10^6}{1+\dfrac{2.17\times10^9\times0.61}{1.9\times10^{11}\times0.005}}}=952.2\ \text{m/s}
$$
$$
\frac{2L}{c}=\frac{6400}{952.2}=6.72\ \text{s}
$$

#### Step 3: Hammer pressure
Static pressure: $p_s=\rho g H=1000\times9.81\times61=0.598$ MPa.

**Closure in 4 s** ($<6.72$ s, rapid):
$$
\Delta p=\rho cV=1000\times952.2\times4.30=4.09\ \text{MPa}\ (417\ \text{m})
$$

**Closure in 60 s** ($>6.72$ s, gradual):
$$
\Delta p=\frac{\rho LV}{t_c}=\frac{1000\times3200\times4.30}{60}=0.229\ \text{MPa}\ (23.4\ \text{m})
$$

#### Step 4: Stress in the pipe wall, $\sigma=pD/2t$
| Case | Total pressure $p_s+\Delta p$ (MPa) | Hoop stress (MPa) |
|---|---|---|
| Static only | 0.598 | 36.5 |
| 4 s closure | 4.69 | 286.3 |
| 60 s closure | 0.828 | 50.5 |

#### Step 5: Minimum safe closure time
Working stress $=16000$ psi $=16000\times6894.76=110.3$ MPa. Allowable pressure:
$$
p_{all}=\frac{2t\sigma}{D}=\frac{2\times0.005\times110.3}{0.61}=1.808\ \text{MPa}
$$
Allowable hammer rise $=p_{all}-p_s=1.808-0.598=1.210$ MPa. Using the gradual-closure formula:
$$
t_{min}=\frac{\rho LV}{\Delta p_{all}}=\frac{1000\times3200\times4.30}{1210051}=11.4\ \text{s}
$$
This is greater than $2L/c=6.72$ s, so the gradual-closure formula is valid.

**Answer: hammer pressure = 4.09 MPa for 4 s closure and 0.229 MPa for 60 s closure; wall stress = 286 MPa and 50.5 MPa respectively; minimum safe closure time ≈ 11.4 s.**

The 4 s closure gives a stress far above the working stress, so it is unsafe.

## Check yourself

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

1. How is the initial velocity found in this problem?
   A. From the head loss: H = (1 + fL/D) V^2/2g with f from Colebrook
   B. From the Joukowsky equation
   C. From the wall thickness
   D. Directly from V = c/g
   Answer: A. The pipe discharges under head 61 m, so the head is balanced by friction plus exit velocity head. (Key point: Initial velocity)

2. With 2L/c = 6.72 s, which closures are rapid?
   A. 4 s only
   B. Both 4 s and 60 s
   C. 60 s only
   D. Neither
   Answer: A. 4 s < 6.72 s is rapid (rho c V); 60 s is gradual (rho L V / t_c). (Key point: Rapid or gradual)

3. Why is the hammer pressure for 60 s much smaller (0.229 MPa vs 4.09 MPa)?
   A. Slow closure spreads the velocity change over a long time, so dp = rho L V / t_c is small
   B. The pipe is thicker for 60 s
   C. The wave speed is zero
   D. Water density falls
   Answer: A. The pressure rise is inversely proportional to closing time for gradual closure. (Key point: Effect of slow closure)

4. How is the minimum safe closure time obtained?
   A. From the stress at 4 s only
   B. Allowable pressure from working stress, minus static pressure, then t = rho L V / dp_allow
   C. t = 2L/c always
   D. t = D/V
   Answer: B. p_allow = 2t sigma / D, allowable hammer = p_allow - p_s, giving about 11.4 s. (Key point: Minimum safe closure time)

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