---
title: "Pressure rise for instantaneous closure, elastic pipe"
subject: "Hydraulics"
chapter: 4
key: "semester_4/hydraulics:elastic-pipe-pressure-rise-derivation"
source: https://www.bctengineer.com/q/semester_4/hydraulics/elastic-pipe-pressure-rise-derivation
site: BCT Engineering
---

# Pressure rise for instantaneous closure, elastic pipe

Hydraulics · Chapter 4: Unsteady flow in pipes

- Asked in: 2069 Bhadra, 7+1 marks
- Video: [Derivation On Sudden Closure Of Valve When Pipe Is Elastic - Pipe Network and Water Hammer Effect](https://www.youtube.com/watch?v=2iDmOeajflE) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/elastic-pipe-pressure-rise-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-4

## Question

Derive an expression for the pressure rise due to instantaneous closure of valve considering the pipe to be elastic. From the derived expression for elastic pipe, obtain the pressure rise for rigid pipe.

## Short answer

From momentum: Δp = ρcV. Equating kinetic energy to strain energy of water and pipe wall gives c = √[K/ρ / (1+KD/(tE))], hence Δp = V√[ρK/(1+KD/(tE))]. For rigid pipe (E→∞), c = √(K/ρ) and Δp = V√(Kρ).

## Answer

Pressure rise on sudden valve closure is found from the momentum (impulse) equation applied to the pressure wave, and the wave speed from the continuity (energy/volume) equation, including elasticity of the pipe.

#### Set-up
Pipe of diameter $D$, wall thickness $t$, Young's modulus $E$ of pipe, water density $\rho$, bulk modulus $K$, flow velocity $V$ before closure. The valve closes suddenly at the downstream end. A pressure wave moves upstream with celerity $c$, bringing the water to rest and raising pressure by $\Delta p$. Let the wave advance a distance $c\,\Delta t$ in time $\Delta t$.

```text
 Reservoir  <---- wave front ----  Valve
 ======= (V) -> |  V=0, p+dp  | ====|
                  <-- c dt -->
```

#### 1. Momentum (impulse) equation
Mass of water brought to rest in time $\Delta t$: $m=\rho A\,c\,\Delta t$. Change in momentum $=\rho A c\,\Delta t\,(0-V)$. The net force is $\Delta p\,A$ (acting against the flow).
$$
-\Delta p\,A\,\Delta t=-\rho A c\,\Delta t\,V\;\Rightarrow\;\Delta p=\rho cV
$$

#### 2. Wave velocity (strain energy)
Kinetic energy lost $=\tfrac12 m V^2=\tfrac12\rho A c\Delta t\,V^2$.

Strain energy stored in water (volume $A c\Delta t$): $\dfrac{\Delta p^2}{2K}Ac\Delta t$.

Strain energy stored in pipe wall (hoop stress $\sigma=\Delta p D/2t$, strain energy per unit volume $\sigma^2/2E$, wall volume $\pi D t\,c\Delta t$):
$$
\frac{1}{2E}\left(\frac{\Delta pD}{2t}\right)^2\pi D t\,c\Delta t=\frac{\Delta p^2}{2}\,\frac{D}{tE}\,A\,c\Delta t
$$
Energy balance:
$$
\frac12\rho V^2=\frac{\Delta p^2}{2K}+\frac{\Delta p^2}{2}\frac{D}{tE}
$$
Substitute $\Delta p=\rho cV$:
$$
\frac12\rho V^2=\frac{\rho^2c^2V^2}{2}\left(\frac1K+\frac{D}{tE}\right)\;\Rightarrow\;c=\sqrt{\frac{1}{\rho\left(\dfrac1K+\dfrac{D}{tE}\right)}}=\sqrt{\frac{K/\rho}{1+\dfrac{KD}{tE}}}
$$

#### 3. Pressure rise for elastic pipe
$$
\boxed{\Delta p=\rho cV=V\sqrt{\frac{\rho K}{1+\dfrac{KD}{tE}}}}\qquad\text{(head rise }\Delta H=\frac{cV}{g}\text{)}
$$

#### 4. Rigid pipe
A rigid pipe does not stretch: $E\to\infty$, so $KD/tE\to0$ and
$$
c=\sqrt{\frac{K}{\rho}},\qquad \Delta p=\rho cV=V\sqrt{K\rho}
$$
This is the maximum possible pressure rise; it is larger than the value for an elastic pipe.

## Check yourself

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

1. Applying the impulse-momentum equation to the water brought to rest gives
   A. dp = K V
   B. dp = V/c
   C. dp = rho g V
   D. dp = rho c V
   Answer: D. -dp A dt = -rho A c dt V gives dp = rho c V. (Key point: Momentum step)

2. The wave speed is obtained by equating
   A. Kinetic energy lost to strain energy stored in water and pipe wall
   B. Momentum to mass
   C. Weight to pressure
   D. Friction to velocity head
   Answer: A. 1/2 rho V^2 = dp^2/2K + dp^2 D/2tE. (Key point: Energy balance)

3. The elastic wave speed is
   A. K/(rho D)
   B. sqrt(E/rho)
   C. sqrt(K/rho)(1 + KD/tE)
   D. sqrt((K/rho)/(1 + KD/(tE)))
   Answer: D. Substituting dp = rho c V yields this expression. (Key point: Wave speed result)

4. For a rigid pipe, E tends to infinity, so
   A. c becomes zero
   B. KD/tE tends to 0 and c = sqrt(K/rho)
   C. dp becomes zero
   D. c = sqrt(E/rho)
   Answer: B. This gives the maximum possible rise, dp = V sqrt(K rho). (Key point: Rigid limit)

## Related in Hydraulics

- [Pressure rise: rigid versus elastic pipe](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Arigid-vs-elastic-pipe-pressure-rise-5km) — Hydraulics, chapter 4
- [Pipe wall stress for 3 s valve closure](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Avalve-closure-3s-stress-in-pipe-wall) — Hydraulics, chapter 4
- [Pressure variation after sudden valve closure](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Awater-hammer-pressure-time-valve-and-midpoint) — Hydraulics, chapter 4
- [Penstock pressure rise and pressure-time sketch](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Asurge-tank-penstock-pressure-rise) — Hydraulics, chapter 4
- [Gradual valve closure with retardation law](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Agradual-valve-closure-head-at-valve) — Hydraulics, chapter 4
- [Rapid closure time and peak hammer pressure](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acast-iron-pipe-rapid-closure-time-and-length) — Hydraulics, chapter 4

## Related in other subjects

- [Pressure rise for gradual and sudden closure](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Awater-hammer-valve-closure) — Hydro Power, chapter 1
- [Water hammer in elastic cast-iron pipe](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Awater-hammer-cast-iron-pipe) — Hydro Power, chapter 1