---
title: "Surge tank: tunnel 5 m dia, 28 cumec"
subject: "Hydropower Engineering"
chapter: 6
key: "semester_7/hydropower_engineering:surge-numerical-tunnel-5m-28cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/surge-numerical-tunnel-5m-28cumec
site: BCT Engineering
---

# Surge tank: tunnel 5 m dia, 28 cumec

Hydropower Engineering · Chapter 6: Water Conveyance Structures

- Asked in: 2081 Baishakh, 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-tunnel-5m-28cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6

## Question

A hydropower electrical project has a concrete lined tunnel of 5.0 m diameter operating under a gross head of 200 m. Discharge through the tunnel is 28 cumec and it has a surge tank of 300 m² at the end of the tunnel. Head loss due to friction under steady state condition is 2.5% of gross head. Assume the friction factor of the tunnel to be 0.015. Find the total length of the tunnel, the maximum upsurge and downsurge in the tank, and calculate the factor of safety of the surge tank.

## Short answer

The total tunnel length is 16,080 m. Maximum upsurge is 11.64 m above reservoir level, downsurge is 8.57 m below, and the surge tank factor of safety against instability is 8.94.

## Answer

#### Given and tunnel length
$A_t = \frac{\pi}{4}(5)^2 = 19.635\ \text{m}^2$, $V = 28/19.635 = 1.426$ m/s, $A_s = 300\ \text{m}^2$, $h_f = 0.025 \times 200 = 5.0$ m.

$$
L = \frac{h_f\,2gD}{fV^2} = \frac{5.0 \times 2(9.81)(5)}{0.015 \times 1.426^2} = 16{,}080\ \text{m}
$$

**Total tunnel length $\approx 16.1$ km.**

#### Maximum upsurge and downsurge
Frictionless amplitude: $Z_0 = 1.426\sqrt{\dfrac{16080 \times 19.635}{9.81 \times 300}} = 14.77$ m.

With friction ($h_f = 5$ m), the upsurge for full load rejection (measured above reservoir level) is found from the equations of motion and continuity:

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

Solving by trial gives $Z_{max} = 11.64$ m above reservoir level. Integrating the reversed-flow equations step by step, the first downswing after the upsurge reaches $8.57$ m below reservoir level.

| Quantity | Value |
|---|---|
| Frictionless amplitude $Z_0$ | 14.77 m |
| Maximum upsurge | 11.64 m above reservoir level |
| Maximum downsurge | 8.57 m below reservoir level |

#### Factor of safety (Thoma stability)
Head-loss coefficient $\alpha = h_f/V^2 = 5/1.426^2 = 2.459\ \text{s}^2/\text{m}$. Net head $H - h_f = 195$ m.

$$
A_{th} = \frac{L A_t}{2g\alpha (H-h_f)} = \frac{16080 \times 19.635}{2(9.81)(2.459)(195)} = 33.56\ \text{m}^2
$$

$$
\text{FOS} = \frac{A_s}{A_{th}} = \frac{300}{33.56} = 8.94
$$

**Answer: $L \approx 16{,}080$ m; upsurge $= 11.64$ m; downsurge $= 8.57$ m; factor of safety $= 8.94$ (the tank is very stable, since $A_s \gg A_{th}$).**

## Check yourself

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

1. How is L obtained from the friction loss?
   A. L = hf f D
   B. L = hf/V
   C. L = V^2/2g
   D. L = hf 2gD/(f V^2) = 16,080 m
   Answer: D. From hf = fLV^2/(2gD), hf = 5 m. (Key point: Tunnel length)

2. The frictionless amplitude Z0 is?
   A. 11.64 m
   B. 14.77 m
   C. 5 m
   D. 8.57 m
   Answer: B. V sqrt(L At/(g As)) = 1.426 x 10.36. (Key point: Frictionless amplitude)

3. Compared with Z0 = 14.77 m, the upsurge with friction is?
   A. Equal
   B. Larger, 17 m
   C. Zero
   D. Smaller, 11.64 m
   Answer: D. Friction damps the surge. (Key point: Friction effect)

4. How is the factor of safety computed?
   A. A_s/A_th = 300/33.56 = 8.94
   B. A_s x A_th
   C. A_th/A_s
   D. Z0/Zmax
   Answer: A. A tank area well above Thoma area is stable. (Key point: Factor of safety)

## Related in Hydropower Engineering

- [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 tank area, up/downsurge, invert level](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-tunnel-5m-4000m-pressure-shaft) — 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
- [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×
- [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 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×

## Related in other subjects

- [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
- [Velocity, head loss, discharge in pipes BC, CD, BF, FD (0.1 m)](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Aloop-c72) — Water Supply Engineering, chapter 7
- [Design: 300 m head plant, GCB rating](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2081-300m-gcb) — Power Plant Design, chapter 4, asked 11×
- [Discharge and head loss in horizontal pipe](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apipe-discharge-head-loss) — Hydro Power, chapter 1