---
title: "Surge numerical: 4.5 m tunnel, 2500 m, 25 m³/s"
subject: "Hydropower Engineering"
chapter: 6
key: "semester_7/hydropower_engineering:surge-numerical-tunnel-4-5m-2500m-25cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/surge-numerical-tunnel-4-5m-2500m-25cumec
site: BCT Engineering
---

# Surge numerical: 4.5 m tunnel, 2500 m, 25 m³/s

Hydropower Engineering · Chapter 6: Water Conveyance Structures

- Asked in: 2072 Kartik, 8 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-tunnel-4-5m-2500m-25cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6

## Question

In a hydropower project the headrace tunnel of 4.5 m diameter and 2,500 m length carries 25 m³/s discharge to the surge tank of 10 m diameter. The penstock from the surge tank to the power house has 3.5 m diameter and 1000 m length. Considering the case of instantaneous closure, find the maximum height of the surge tank required and the time period of oscillation of wave. Assume friction factor = 0.02.

## Short answer

Tunnel velocity 1.572 m/s, friction loss 1.40 m and frictionless amplitude 11.29 m. With instantaneous closure the maximum upsurge is 10.38 m above reservoir level; first downswing is 9.01 m below. Oscillation period is 222.9 s, about 3.7 min.

## 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
$A_t = \frac{\pi}{4}(4.5)^2 = 15.904\ \text{m}^2$, $A_s = \frac{\pi}{4}(10)^2 = 78.54\ \text{m}^2$, $L = 2500$ m, $f = 0.02$.

- $V = 25/15.904 = 1.572$ m/s
- $h_f = \dfrac{0.02 \times 2500 \times 1.572^2}{2(9.81)(4.5)} = 1.40$ m
- $Z_0 = 1.572\sqrt{\dfrac{2500 \times 15.904}{9.81 \times 78.54}} = 11.29$ m

Instantaneous closure means full rejection, so the water rises in the tank. The penstock is downstream of the tank and does not change the tank oscillation.

#### Maximum height of surge tank
With friction, $Z_{max} = 10.38$ m above reservoir level (frictionless value 11.29 m). The tank should rise at least 10.4 m above the reservoir level, plus freeboard of about 1 to 2 m. The first downswing is 9.01 m below reservoir level.

#### Time period

$$
T = 2\pi\sqrt{\frac{2500 \times 78.54}{9.81 \times 15.904}} = 222.9\ \text{s} \approx 3.7\ \text{min}
$$

**Answer: maximum upsurge height $= 10.38$ m above reservoir level; $T = 222.9$ s.**

## Check yourself

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

1. What is the steady friction loss hf in the 4.5 m tunnel?
   A. About 1.40 m
   B. About 11.29 m
   C. About 0.14 m
   D. About 10.38 m
   Answer: A. hf = f L V^2/(2gD) = 0.02 x 2500 x 1.572^2 / (2 x 9.81 x 4.5) = 1.40 m. (Key point: Friction head loss)

2. What does Z0 = 11.29 m represent?
   A. The downsurge below reservoir level
   B. Upsurge neglecting friction, an upper bound; the actual Zmax with friction is 10.38 m
   C. The water hammer head
   D. The height of the tank freeboard
   Answer: B. Z0 ignores friction so Zmax with friction is slightly lower. (Key point: Frictionless amplitude)

3. How does the 1000 m, 3.5 m penstock affect the tank oscillation?
   A. It doubles the period
   B. It adds its friction to hf
   C. It does not, since it lies downstream of the tank
   D. It sets the tank diameter
   Answer: C. Only the tunnel upstream of the tank governs the mass oscillation. (Key point: Penstock role)

4. What is the period of oscillation?
   A. T = 11.3 s
   B. T = 445.8 s
   C. T = 1.4 s
   D. T = 222.9 s (about 3.7 min)
   Answer: D. T = 2 pi sqrt(L As/(g At)) = 2 pi sqrt(2500 x 78.54 /(9.81 x 15.904)) = 222.9 s. (Key point: Period)

## Close variants set in other exams

- [Surge numerical: 4 m tunnel, 4000 m, 20 m³/s](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-tunnel-4m-4000m-20cumec) — Hydropower Engineering, chapter 6, asked 2×

## Related in Hydropower Engineering

- [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 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×
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- [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 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 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×

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