---
title: "Surge tank: 7.5 m³/s through 2.5 m tunnel"
subject: "Hydropower Engineering"
chapter: 6
key: "semester_7/hydropower_engineering:surge-numerical-7-5cumec-2-5m"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/surge-numerical-7-5cumec-2-5m
site: BCT Engineering
---

# Surge tank: 7.5 m³/s through 2.5 m tunnel

Hydropower Engineering · Chapter 6: Water Conveyance Structures

- Asked in: 2082 Bhadra, 6 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-7-5cumec-2-5m
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6

## Question

A hydropower project is designed to carry a flow of 7.5 m³/sec through a tunnel of diameter 2.5 m. A simple surge tank of diameter 7 m is located at a distance of 2,500 m from the reservoir. Calculate the total time period of oscillation of wave for full load rejection. Also, estimate the height of the surge tank. Take friction factor = 0.018.

## Short answer

The oscillation time period is 280.8 s. Maximum upsurge is 7.35 m above reservoir level, and the surge tank height required is about 15 m, including freeboard.

## 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: $T = 2\pi\sqrt{\dfrac{L A_s}{g A_t}}$.

#### Calculation
$A_t = \frac{\pi}{4}(2.5)^2 = 4.909\ \text{m}^2$, $A_s = \frac{\pi}{4}(7)^2 = 38.48\ \text{m}^2$, $L = 2500$ m, $f=0.018$.

- $V = 7.5/4.909 = 1.528$ m/s
- $h_f = \dfrac{0.018 \times 2500 \times 1.528^2}{2(9.81)(2.5)} = 2.14$ m
- $Z_0 = 1.528\sqrt{\dfrac{2500 \times 4.909}{9.81 \times 38.48}} = 8.71$ m

#### Time period of oscillation

$$
T = 2\pi\sqrt{\frac{2500 \times 38.48}{9.81 \times 4.909}} = 280.8\ \text{s} \approx 4.7\ \text{min}
$$

#### Height of surge tank
Solving the upsurge equation with friction: $Z_{max} = 7.35$ m above reservoir level. The first downsurge is $5.75$ m below reservoir level, so the water level swings through about $13.1$ m. Adding a freeboard of about 2 m at the top, the surge tank must be at least about $15$ m high above the minimum level.

**Answer: $T = 280.8$ s; maximum upsurge $= 7.35$ m above reservoir level (tank height about 15 m including range and freeboard).**

## Check yourself

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

1. The period T is found from?
   A. L/V
   B. 2 pi sqrt(L A_s/(g A_t)) = 280.8 s
   C. 2 pi sqrt(L A_t/(g A_s))
   D. sqrt(Z0)
   Answer: B. A_s = 38.48, A_t = 4.909. (Key point: Period)

2. Tunnel velocity is?
   A. 0.76 m/s
   B. 3.05 m/s
   C. 1.528 m/s
   D. 7.5 m/s
   Answer: C. 7.5/4.909 = 1.528 m/s. (Key point: Velocity)

3. How is tank height estimated?
   A. Head loss only
   B. Tunnel length/100
   C. Z0 only
   D. Upsurge + downsurge range plus freeboard, about 15 m
   Answer: D. 7.35 + 5.75 = 13.1 m plus about 2 m freeboard. (Key point: Tank height)

4. The steady friction loss hf is about?
   A. 8.71 m
   B. 2.14 m
   C. 15 m
   D. 7.35 m
   Answer: B. hf = fLV^2/(2gD). (Key point: Friction loss)

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