---
title: "Penstock pressure rise and pressure-time sketch"
subject: "Hydraulics"
chapter: 4
key: "semester_4/hydraulics:surge-tank-penstock-pressure-rise"
source: https://www.bctengineer.com/q/semester_4/hydraulics/surge-tank-penstock-pressure-rise
site: BCT Engineering
---

# Penstock pressure rise and pressure-time sketch

Hydraulics · Chapter 4: Unsteady flow in pipes

- Asked in: 2079 Chaitra, 6 marks
- Video: [Show that the rise in pressure head due to the valve closure is H=LV/GT | Hydraulics | Prashant YT |](https://www.youtube.com/watch?v=zjydlqES7Uo) — Prashant YT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/surge-tank-penstock-pressure-rise
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-4

## Question

Water is flowing at 4 m/s in a penstock 4500 m long. If celerity of the pressure wave travelling in the pipe due to the sudden closure of a valve at the downstream end is given as 1500 m/s, what will be the maximum pressure rise? Show how the pressure changes with time at the 400 m upstream of penstock length from the valve.

## Short answer

Maximum pressure rise = 612 m of water (6.0 MPa). At 400 m upstream of valve: head elevated by 612 m from 0.27 to 5.73 s, normal from 5.73 to 6.27 s, depressed by 612 m from 6.27 to 11.73 s, then normal; cycle repeats every 12 s.

## Answer

**Data.** $V = 4$ m/s, $L = 4500$ m, $c = 1500$ m/s, sudden valve closure at the downstream end.

#### Maximum pressure rise (Joukowsky)
Sudden closure means $t_c \le 2L/c = 6.0$ s (rapid closure), so the full rise occurs:
$$
\Delta H = \frac{cV}{g} = \frac{1500(4)}{9.81} = 611.6\ \text{m of water}
$$
$$
\Delta p = \rho cV = 1000(1500)(4) = 6.0\times10^{6}\ \text{Pa} = 6.0\ \text{MPa}
$$

#### Pressure-time variation at 400 m upstream of the valve
Distance from the valve $d = 400$ m. Wave arrival times:
- $t_1 = \dfrac{d}{c} = \dfrac{400}{1500} = 0.267$ s: the pressure wave arrives; pressure rises by $+611.6$ m ($+6.0$ MPa)
- $t_2 = \dfrac{2L - d}{c} = 5.733$ s: the negative wave reflected from the reservoir arrives; pressure returns to normal
- $t_3 = \dfrac{2L + d}{c} = 6.267$ s: the negative wave reflected from the valve arrives; pressure drops by $611.6$ m
- $t_4 = \dfrac{4L - d}{c} = 11.733$ s: the positive wave reflected from the reservoir arrives; pressure returns to normal
- The cycle repeats every $4L/c = 12.0$ s.

| Time (s) | Pressure change at 400 m |
|---|---|
| 0 to 0.27 | 0 |
| 0.27 to 5.73 | $+612$ m |
| 5.73 to 6.27 | 0 |
| 6.27 to 11.73 | $-612$ m |
| 11.73 to 12 | 0 |

```text
  *********************


**---------------------***---------------------**


                          *********************
(top = 1.0, bottom = -1.0; '-' is zero line)
```
(+1 = $+\Delta H$, $-1$ = $-\Delta H$ relative to the normal pressure; time from 0 to $4L/c = 12$ s.)

**Answer:** Maximum pressure rise $= 612$ m of water ($6.0$ MPa). At 400 m from the valve the head is raised by 612 m from 0.267 s to 5.733 s and lowered by 612 m from 6.267 s to 11.733 s; this stepped wave repeats every 12 s.

## Check yourself

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

1. Why does the full Joukowsky rise apply for sudden closure with L = 4500 m, c = 1500 m/s?
   A. The closure time is not more than 2L/c = 6.0 s
   B. The closure time must exceed 6 s
   C. Because velocity is above 3 m/s
   D. Because 4L/c = 12 s
   Answer: A. Closure within 2L/c is rapid, so the full pressure rise is produced. (Key point: Rapid closure check)

2. What is the maximum pressure rise (V = 4 m/s, c = 1500 m/s)?
   A. About 61 m (0.6 MPa)
   B. About 612 m (6.0 MPa)
   C. About 1224 m (12 MPa)
   D. About 153 m (1.5 MPa)
   Answer: B. delta H = cV/g = 1500 x 4 / 9.81 = 611.6 m, i.e. rho c V = 6.0 MPa. (Key point: Maximum pressure rise)

3. When does the positive pressure wave first arrive at a point 400 m from the valve?
   A. 0.267 s
   B. 5.733 s
   C. 6.267 s
   D. 3.0 s
   Answer: A. t1 = d/c = 400/1500 = 0.267 s; the reservoir reflection cancels it at (2L-d)/c = 5.733 s. (Key point: Timing at 400 m from the valve)

## Related in Hydraulics

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