---
title: "Penstock thickness with gate closure time"
subject: "Hydropower Engineering"
chapter: 6
key: "semester_7/hydropower_engineering:penstock-thickness-water-hammer-600m"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/penstock-thickness-water-hammer-600m
site: BCT Engineering
---

# Penstock thickness with gate closure time

Hydropower Engineering · Chapter 6: Water Conveyance Structures

- Asked in: 2074 Ashwin, 8 marks
- Video: [Design of Surge Tank(Shaft) | Water Hammer Effect | Hydropower Engineering | IOE , TU , PU](https://www.youtube.com/watch?v=r1J4tINkknw) — Parash Joshi - Civil Construction and Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/penstock-thickness-water-hammer-600m
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6

## Question

A hydropower plant has planned to use a steel penstock pipe of length 600 m having a diameter of 0.8 m to carry a discharge of 5 m³/s. The static head available is 80 m. The wave velocity, design stress and joint efficiency for the penstock pipe are 1200 m/s, 1326 kg/cm² and 85% respectively. What thickness of the penstock pipe would you recommend for the power plant if the gate closure time is 30 seconds?

## Short answer

Closure is slow because Tc=30 s >> 2L/a=1 s. Water-hammer rise ΔH = 2Lv/(gTc) = 40.56 m, so total head 120.56 m, pressure 12.06 kg/cm²; t = 12.06×80/(2×1326×0.85) = 4.28 mm. Recommend 6 mm plate with corrosion allowance.

## Answer

#### Data
$L = 600$ m, $D = 0.8$ m $= 80$ cm, $Q = 5\ \text{m}^3/\text{s}$, static head $H = 80$ m, $a = 1200$ m/s, $\sigma = 1326\ \text{kg/cm}^2$, $\eta_j = 0.85$, closure time $T_c = 30$ s.

#### Type of closure
Velocity: $v = Q/A = 5/(\frac{\pi}{4}0.8^2) = 9.947$ m/s.

Critical time $2L/a = 2(600)/1200 = 1$ s. Since $T_c = 30\ \text{s} \gg 1$ s, the closure is slow, so the Joukowsky value $av/g$ does not apply; use the Michaud (Allievi) formula.

#### Water hammer rise

$$
\Delta H = \frac{2Lv}{gT_c} = \frac{2 \times 600 \times 9.947}{9.81 \times 30} = 40.56\ \text{m}
$$

#### Design pressure and thickness
$$
H_{total} = 80 + 40.56 = 120.56\ \text{m} \;\Rightarrow\; p = 12.06\ \text{kg/cm}^2
$$

$$
t = \frac{pD}{2\sigma\eta_j} = \frac{12.06 \times 80}{2 \times 1326 \times 0.85} = 0.428\ \text{cm} = 4.28\ \text{mm}
$$

Practical checks: minimum thickness for handling, $t_{min} = \dfrac{D(\text{mm}) + 800}{400} = \dfrac{800+800}{400} = 4$ mm. Allow 1.5 to 2 mm corrosion allowance.

**Answer: calculated $t = 4.3$ mm; recommend a 6 mm thick steel penstock (4.3 mm plus corrosion allowance, rounded to a standard plate).**

## Check yourself

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

1. Why is the Joukowsky formula not used here?
   A. Closure time 30 s is much longer than 2L/a = 1 s, so closure is slow
   B. Closure time is shorter than 2L/a
   C. The wave velocity is unknown
   D. The static head is too high
   Answer: A. Rapid closure (T <= 2L/a) uses a v/g; slow closure gives a much smaller rise. (Key point: Closure type)

2. Which formula gives the water hammer rise for slow closure?
   A. dH = a v / g = 1216 m
   B. dH = 2 L v / (g Tc) = 40.56 m
   C. dH = v^2/2g
   D. dH = L v / a
   Answer: B. Michaud/Allievi formula applies for slow closure. (Key point: Slow closure formula)

3. What total head is used for thickness?
   A. 80 m only
   B. 1216 m
   C. Static head + rise = 80 + 40.56 = 120.56 m
   D. 40.56 m only
   Answer: C. Thickness is designed for static plus water hammer head. (Key point: Design head)

4. What is calculated thickness and recommendation?
   A. 43 mm
   B. 0.43 mm
   C. 12 mm
   D. 4.3 mm calculated, recommend about 6 mm with corrosion allowance and standard plate
   Answer: D. t = 12.06 x 80/(2 x 1326 x 0.85) = 0.428 cm; add corrosion allowance, round to 6 mm. (Key point: Final thickness)

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