---
title: "Surge numerical: 13 m tank, 1.8 km tunnel"
subject: "Hydropower Engineering"
chapter: 6
key: "semester_7/hydropower_engineering:surge-numerical-13m-tank-1-8km-tunnel"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/surge-numerical-13m-tank-1-8km-tunnel
site: BCT Engineering
---

# Surge numerical: 13 m tank, 1.8 km tunnel

Hydropower Engineering · Chapter 6: Water Conveyance Structures

- Asked in: 2079 Bhadra, 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-13m-tank-1-8km-tunnel
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6

## Question

A RoR hydel plant has a circular surge tank of 13 m diameter at the end of a 1.8 km long headrace pressure tunnel with 3.95 m diameter. The penstock system consists of 4 numbers, 400 m long, 1.30 m diameter each. Calculate the maximum up-surge, down-surge and time of oscillations if the frictional factor for tunnel and penstock are 0.016 and 0.025 respectively.

## Short answer

Assuming design discharge 30 m3/s, tunnel velocity 2.448 m/s gives friction loss 2.23 m. Maximum upsurge is 8.65 m above reservoir level, downsurge 6.90 m below; period of oscillation is 280.1 s, about 4.7 min. For other discharges surge heights scale roughly with Q.

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

#### Assumption
The discharge is not given. Assume a design discharge $Q = 30\ \text{m}^3/\text{s}$ (4 penstocks of 7.5 m³/s, about 5.65 m/s each). Surge heights scale almost in proportion to $V$, so for another $Q$ repeat the same steps.

#### Calculation
$A_t = \frac{\pi}{4}(3.95)^2 = 12.25\ \text{m}^2$, $A_s = \frac{\pi}{4}(13)^2 = 132.73\ \text{m}^2$, $L = 1800$ m, $f = 0.016$.

- $V = 30/12.25 = 2.448$ m/s
- $h_f = \dfrac{0.016 \times 1800 \times 2.448^2}{2(9.81)(3.95)} = 2.23$ m
- $Z_0 = 2.448\sqrt{\dfrac{1800 \times 12.25}{9.81 \times 132.73}} = 10.08$ m

The penstock friction (0.025) acts downstream of the tank, so it does not enter the tank oscillation.

#### Results (full load rejection)
- **Maximum up-surge** $= 8.65$ m above reservoir level.
- **Maximum down-surge** $= 6.90$ m below reservoir level.
- **Time of oscillation**

$$
T = 2\pi\sqrt{\frac{1800 \times 132.73}{9.81 \times 12.25}} = 280.1\ \text{s} \approx 4.7\ \text{min}
$$

**Answer (for $Q = 30$ m³/s): up-surge 8.65 m; down-surge 6.90 m; $T = 280.1$ s. For other discharges the surges grow roughly in proportion to $Q$ (e.g. $Q = 20$ m³/s: 6.07 m up, 5.16 m down; $Q = 40$ m³/s: 10.94 m up, 8.27 m down).**

## Check yourself

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

1. Why does the penstock friction factor 0.025 not enter the surge tank oscillation calculation?
   A. The penstock lies downstream of the tank, so its flow is not part of the tank-tunnel oscillation
   B. Because the penstock is shorter than the tunnel
   C. Because penstock friction is always zero
   D. Because four penstocks cancel each other
   Answer: A. The tank oscillation involves the water in the tunnel between reservoir and tank; the penstock is beyond the tank. (Key point: Penstock friction irrelevant)

2. The discharge is not given. What does the answer do about it?
   A. Takes Q from the penstock friction factor
   B. Assumes Q = 30 m3/s and notes the surges scale roughly in proportion to Q
   C. Sets Q equal to zero
   D. Computes Q from the tank diameter alone
   Answer: B. It assumes 30 m3/s and shows that surge heights scale nearly with velocity, hence with Q. (Key point: Missing discharge assumption)

3. For this tank, what are Z0 and the period T?
   A. Z0 = 2.23 m and T = 280.1 s
   B. Z0 = 10.08 m and T = 5.3 s
   C. Z0 = 10.08 m and T = 280.1 s
   D. Z0 = 8.65 m and T = 317 s
   Answer: C. Z0 = V sqrt(L At/(g As)) = 10.08 m; T = 2 pi sqrt(L As/(g At)) = 280.1 s. (Key point: Frictionless amplitude and period)

4. Why is the maximum upsurge (8.65 m) less than Z0 = 10.08 m?
   A. The tank is too large
   B. Penstock friction opposes it
   C. The discharge was reduced by half
   D. Tunnel friction (hf = 2.23 m) dissipates energy
   Answer: D. Friction in the tunnel reduces the amplitude below the frictionless value. (Key point: Effect of friction on surge)

## 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: 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×
- [Surge numerical: 4.5 m tunnel, 2500 m, 25 m³/s](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-tunnel-4-5m-2500m-25cumec) — Hydropower Engineering, chapter 6, asked 2×
- [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
- [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
- [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×

## 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
- [450 km 400 kV line, 420 MW, 0.95 pf](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A450km-400kv-420mw-095) — Power System Analysis I, chapter 5
- [Water hammer in penstock and surge tank](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Awater-hammer-surge-tank) — Power Plant Equipment, chapter 1, asked 2×
- [Surge impedance and travel time of 400 km line](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Asurge-impedance-220kv-400km) — High Voltage Engineering, chapter 3