---
title: "Peak pressure for 4.5 s closure"
subject: "Hydraulics"
chapter: 4
key: "semester_4/hydraulics:valve-closed-4-5s-peak-pressure-3200m"
source: https://www.bctengineer.com/q/semester_4/hydraulics/valve-closed-4-5s-peak-pressure-3200m
site: BCT Engineering
---

# Peak pressure for 4.5 s closure

Hydraulics · Chapter 4: Unsteady flow in pipes

- Asked in: 2070 Magh, 5 marks
- Video: [Peak Transient Pressure due to Valve Closure](https://www.youtube.com/watch?v=Xls-AI12CvM) — Simple Engineering Snippets
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/valve-closed-4-5s-peak-pressure-3200m
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-4

## Question

A valve is closed in 4.5 s at the down stream end of a 3200 m pipeline carrying water at 2.7 m/s. What is the peak pressure developed by the closure, if the wave travels with velocity of 1000 m/s? Determine the length of pipe subjected to the peak discharge.

## Short answer

Closure time 4.5 s is less than 2L/c = 6.4 s, so rapid closure applies. Peak pressure rise Δp = ρcV = 2.70 MPa (275.2 m head). The peak pressure acts over the first 950 m from the valve.

## Answer

The wave from the valve travels to the reservoir and back in time $2L/c$. If the closure finishes before the reflected wave returns to the valve, the full Joukowsky pressure develops at the valve.

#### Given data
$L=3200$ m, $V=2.7$ m/s, $c=1000$ m/s, $t_c=4.5$ s.

#### Peak pressure
$$
\frac{2L}{c}=\frac{2\times3200}{1000}=6.4\ \text{s}>t_c=4.5\ \text{s}
$$
The closure is rapid (the valve is completely shut before the negative wave returns), so
$$
\Delta p=\rho cV=1000\times1000\times2.7=2.70\ \text{MPa}\quad\left(\Delta H=\frac{cV}{g}=275.2\ \text{m}\right)
$$

#### Length of pipe subjected to the peak pressure
The last pressure increment leaves the valve at $t=t_c$ and reaches a point $x$ from the valve at $t_c+x/c$. The negative wave reflected from the reservoir reaches the same point at $(2L-x)/c$. The full peak is felt only where the last increment arrives first:
$$
t_c+\frac{x}{c}\le\frac{2L-x}{c}\;\Rightarrow\;x\le L-\frac{ct_c}{2}=3200-\frac{1000\times4.5}{2}=950\ \text{m}
$$

```text
 Reservoir                              Valve
 |<------ 2250 m ------>|<---- 950 m --->|
   reduced pressure        peak pressure
```
**Answer: peak pressure rise = 2.70 MPa (275.2 m of water); length of pipe subjected to the peak pressure = 950 m from the valve.**

## Check yourself

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

1. Here 2L/c = 6.4 s and t_c = 4.5 s. The closure is
   A. Instantaneous
   B. Cannot be decided
   C. Rapid
   D. Gradual
   Answer: C. t_c < 2L/c so the valve shuts before the negative wave returns. (Key point: Check of closure)

2. What is the peak pressure rise?
   A. 2.70 MPa
   B. 0.77 MPa
   C. 27 MPa
   D. 6.0 MPa
   Answer: A. dp = rho c V = 1000 x 1000 x 2.7 = 2.70 MPa. (Key point: Peak pressure)

3. The length subjected to peak pressure is
   A. L - c t_c/2 = 950 m
   B. L/2 = 1600 m
   C. c t_c = 4500 m
   D. 2L/c = 6400 m
   Answer: A. 3200 - 1000 x 4.5/2 = 950 m from the valve. (Key point: Length at peak)

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