---
title: "Surge and water hammer: 60 m³/s, 30 m surge tank"
subject: "Hydropower Engineering"
chapter: 6
key: "semester_7/hydropower_engineering:surge-numerical-60cumec-3-penstocks-30m-tank"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/surge-numerical-60cumec-3-penstocks-30m-tank
site: BCT Engineering
---

# Surge and water hammer: 60 m³/s, 30 m surge tank

Hydropower Engineering · Chapter 6: Water Conveyance Structures

- Asked in: 2070 Ashad, 4+2+2 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-60cumec-3-penstocks-30m-tank
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6

## Question

The design discharge through the tunnel of a hydropower project is 60 m³/s and is conveyed by three penstocks to the turbine of 2 m diameter each. Take the length of the tunnel as 7 km, diameter of tunnel is 10 m, friction factor of tunnel is 0.016, friction factor of penstock = 0.04 and velocity of wave in penstock = 1800 m/sec. If a surge tank of 30 m diameter has been provided at the end of the tunnel, find the following: (i) maximum up-surge and down-surge in the tank, (ii) water hammer pressure, (iii) time of oscillation of wave.

## Short answer

Tunnel velocity 0.764 m/s gives friction loss 0.333 m and frictionless amplitude 6.80 m. Maximum upsurge is 6.58 m above reservoir level, downsurge 6.19 m below. Per penstock velocity 6.366 m/s gives water hammer head 1168.1 m (11.46 MPa); oscillation period is 503.5 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
$A_t = \frac{\pi}{4}(10)^2 = 78.54\ \text{m}^2$, $A_s = \frac{\pi}{4}(30)^2 = 706.86\ \text{m}^2$, $L = 7000$ m.

- $V = 60/78.54 = 0.764$ m/s
- $h_f = \dfrac{0.016 \times 7000 \times 0.764^2}{2(9.81)(10)} = 0.333$ m
- $Z_0 = 0.764\sqrt{\dfrac{7000 \times 78.54}{9.81 \times 706.86}} = 6.80$ m

#### (i) Maximum up-surge and down-surge
- Up-surge $= 6.58$ m above reservoir level.
- Down-surge (first downswing) $= 6.19$ m below reservoir level.

#### (ii) Water hammer pressure
Discharge per penstock $= 60/3 = 20$ m³/s; $A_p = 3.142\ \text{m}^2$; $v = 20/3.142 = 6.366$ m/s.

$$
h = \frac{a v}{g} = \frac{1800 \times 6.366}{9.81} = 1168.1\ \text{m}
$$

$p = 11.46\ \text{MPa}$ ($\approx 116.8\ \text{kg/cm}^2$), for instantaneous (rapid) closure.

#### (iii) Time of oscillation

$$
T = 2\pi\sqrt{\frac{7000 \times 706.86}{9.81 \times 78.54}} = 503.5\ \text{s} \approx 8.4\ \text{min}
$$

**Answer: up-surge 6.58 m; down-surge 6.19 m; water hammer head 1168.1 m (11.46 MPa); $T = 503.5$ 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. After full load rejection, why is the first downswing smaller than the frictionless amplitude Z0?
   A. The surge tank area increases during downswing
   B. The penstock friction adds extra head
   C. Friction in the tunnel damps the oscillation (6.19 m below vs Z0 = 6.80 m)
   D. Water hammer absorbs the energy
   Answer: C. Tunnel friction dissipates energy so both the upsurge (6.58 m) and downsurge (6.19 m) are below Z0 = 6.80 m. (Key point: Downsurge versus upsurge)

4. For 20 m3/s per penstock with a = 1800 m/s, v = 6.366 m/s. What is the water hammer head?
   A. About 584 m
   B. About 2336 m
   C. About 6.5 m
   D. About 1168 m
   Answer: D. h = a v / g = 1800 x 6.366 / 9.81 = 1168 m. (Key point: Penstock water hammer)

## 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: 45 m³/s, 20 m surge tank](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-45cumec-3-penstocks) — Hydropower Engineering, chapter 6, asked 2×
- [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
- [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
- [Pumped storage plant: power, turbine, setting, surge](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apumped-storage-turbine-surge-numerical) — Hydropower Engineering, chapter 7
- [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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- [Surge tank objectives and economic penstock diameter factors](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asurge-tank-objectives-economic-diameter) — Hydro Power, chapter 5