---
title: "Surge and water hammer: 45 m³/s, 20 m surge tank"
subject: "Hydropower Engineering"
chapter: 6
key: "semester_7/hydropower_engineering:surge-numerical-45cumec-3-penstocks"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/surge-numerical-45cumec-3-penstocks
site: BCT Engineering
---

# Surge and water hammer: 45 m³/s, 20 m surge tank

Hydropower Engineering · Chapter 6: Water Conveyance Structures

- Asked in: 2082 Baishakh, 7 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/surge-numerical-45cumec-3-penstocks
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6

## Question

A hydropower project has a design discharge of 45 m³/s through a tunnel, which is conveyed by three penstocks to the turbines. The tunnel is 4 km long and 8 m in diameter, with a friction factor of 0.016. Each penstock is 500 m long and 2 m in diameter, with a friction factor of 0.04. The velocity of the wave in the penstocks is 1400 m/s. A surge tank with a diameter of 20 m is provided at the end of the tunnel. For a full load rejection, determine the following: a) Maximum up-surge, b) Maximum down-surge, c) Water hammer pressure, d) Time of oscillation of wave.

## Short answer

Maximum upsurge is 7.02 m, downsurge 6.63 m. Water hammer pressure head due to sudden closure is 681.4 m, and the mass oscillation time period is 317.2 s.

## Answer

#### Method used
Let the tunnel have area $A_t$, length $L$, diameter $D$ and friction factor $f$; the tank has area $A_s$.

- Tunnel velocity $V = Q/A_t$; steady friction loss $h_f = \dfrac{fLV^2}{2gD}$.
- Frictionless amplitude: $Z_0 = V\sqrt{\dfrac{L A_t}{g A_s}}$.
- Upsurge with friction (full load rejection), $Z$ measured above reservoir level, from the equation of motion and continuity:

$$
\frac{Z_{max}}{h_f}=\frac{Z_0^2}{2h_f^2}\left[1-e^{-2h_f (Z_{max}+h_f)/Z_0^2}\right]
$$

- The following down-surge is found by integrating the reversed-flow equations step by step (below reservoir level).
- Period of mass oscillation: $T = 2\pi\sqrt{\dfrac{L A_s}{g A_t}}$.

#### Surge in the tunnel
$A_t = \frac{\pi}{4}(8)^2 = 50.27\ \text{m}^2$, $A_s = \frac{\pi}{4}(20)^2 = 314.16\ \text{m}^2$, $L = 4000$ m.

- $V = 45/50.27 = 0.895$ m/s
- $h_f = \dfrac{0.016 \times 4000 \times 0.895^2}{2(9.81)(8)} = 0.327$ m
- $Z_0 = 0.895\sqrt{\dfrac{4000 \times 50.27}{9.81 \times 314.16}} = 7.23$ m

#### a) Maximum up-surge
$Z_{max} = 7.02$ m above reservoir level.

#### b) Maximum down-surge
First downswing $= 6.63$ m below reservoir level.

#### c) Water hammer pressure
Each penstock carries $45/3 = 15$ m³/s. $A_p = \frac{\pi}{4}(2)^2 = 3.142\ \text{m}^2$, so $v = 15/3.142 = 4.775$ m/s.

Full load rejection means the gates close suddenly (closure time $\le 2L/a = 2(500)/1400 = 0.71$ s), so the Joukowsky rise is

$$
h = \frac{a v}{g} = \frac{1400 \times 4.775}{9.81} = 681.4\ \text{m}
$$

$p = \rho g h = 6.69\ \text{MPa}$ ($\approx 68.1\ \text{kg/cm}^2$).

#### d) Time of oscillation

$$
T = 2\pi\sqrt{\frac{4000 \times 314.16}{9.81 \times 50.27}} = 317.2\ \text{s} \approx 5.3\ \text{min}
$$

(The pressure wave in the penstock itself has a period $4L_p/a = 4(500)/1400 = 1.43$ s.)

**Answer: up-surge 7.02 m; down-surge 6.63 m; water hammer head 681.4 m (6.69 MPa); $T = 317.2$ s.**

## Check yourself

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

1. Which expression gives the frictionless surge amplitude Z0 in the tunnel-surge tank system?
   A. Z0 = V sqrt(L At / (g As))
   B. Z0 = V sqrt(L As / (g At))
   C. Z0 = V sqrt(g At / (L As))
   D. Z0 = V^2 / (2g)
   Answer: A. Z0 comes from equating the kinetic energy of the tunnel water to the potential energy gained in the tank, giving V sqrt(L At/(g As)). (Key point: Frictionless amplitude)

2. Which formula gives the period of mass oscillation of the surge tank?
   A. T = 2 pi sqrt(L At / (g As))
   B. T = 2 pi sqrt(L As / (g At))
   C. T = 4 L / a
   D. T = 2 L / a
   Answer: B. The mass oscillation period is 2 pi sqrt(L As/(g At)); 4L/a and 2L/a belong to the pressure wave in the penstock. (Key point: Period of oscillation)

3. How is the penstock velocity v for the water hammer calculation found when three penstocks share 45 m3/s?
   A. Divide 45 m3/s by one penstock area to get 14.3 m/s
   B. Use the tunnel velocity 0.895 m/s
   C. Divide Q by 3 to get 15 m3/s per penstock, then divide by one penstock area, v = 4.775 m/s
   D. Divide 45 m3/s by three times the tunnel area
   Answer: C. Each penstock carries Q/3 = 15 m3/s through its own area of 3.142 m2, so v = 4.775 m/s. (Key point: Penstock velocity for water hammer)

4. Why is the Joukowsky formula h = a v / g used for the water hammer rise after full load rejection?
   A. The closure is slow so friction reduces the rise
   B. The tunnel and the penstock share the same velocity
   C. The surge tank absorbs the pressure wave completely
   D. Gate closure time is not more than 2L/a, so the closure is rapid and the full rise occurs
   Answer: D. For rapid closure (time <= 2L/a) the full Joukowsky rise a v / g develops in the penstock. (Key point: Water hammer closure type)

## Close variants set in other exams

- [Surge and water hammer: 25 m³/s, 15 m surge tank](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-25cumec-2-penstocks-15m-tank) — Hydropower Engineering, chapter 6, asked 3×

## Related in Hydropower Engineering

- [Surge and water hammer: 60 m³/s, 30 m surge tank](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-60cumec-3-penstocks-30m-tank) — Hydropower Engineering, chapter 6, asked 2×
- [Surge numerical: 13 m tank, 1.8 km tunnel](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-13m-tank-1-8km-tunnel) — Hydropower Engineering, chapter 6
- [Surge tank: 7.5 m³/s through 2.5 m tunnel](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-7-5cumec-2-5m) — Hydropower Engineering, chapter 6
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- [Surge tank: tunnel 5 m dia, 28 cumec](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-tunnel-5m-28cumec) — Hydropower Engineering, chapter 6
- [Hydraulic design of surge tank and pressure tunnel](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-tank-hydraulic-design-pressurised-tunnel) — Hydropower Engineering, chapter 6

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